Q: The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be:

(d) 100 m For the same u range, R∝sin(2θ) .So, R2R1=sin(2θ2)sin(2θ1) ⇒R2=50×sin(30)sin(90)=100 m

Question 1

At which point on its path a projectile has the smallest speed?

Accepted Answer

A projectile has the lowest speed at the highest point of its path because the vertical component of velocity is zero at this point.

Question 2

A food packet is dropped from a plane going at an altitude of 100 m. What is the path of the packet as seen from the plane? What is the path as seen from the ground ? If someone asks "what is the actual path", what will you answer?

Accepted Answer

The path of the packet (as seen from the plane) is a vertically downward straight line, as the horizontal velocity of the packet and the plane is the same.As seen from the ground, the path of the packet is a parabola.The path is defined with respect to some reference frame. As there is no absolute reference frame, no actual path is defined.

Question 3

A player hits a baseball at some angle. The ball goes high up in space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?

Accepted Answer

The displacement of the ball and the player is the same, as the initial and final points are the same.

Question 4

In a projectile motion the velocity:

Accepted Answer

(c) Is perpendicular to the acceleration for one instant only.In projectile motion, velocity is perpendicular to acceleration only at the highest point. Here, velocity is along the horizontal direction and acceleration is along the vertically downward direction.

Question 5

Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit he ground first?

Accepted Answer

(c) Both will reach simultaneously because the downward acceleration and the initial velocity in downward direction of the two bullets are the same, they will take the same time to hit the ground and for a half projectile.Time of flight = T =g2h​​ .

Question 6

The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be:

Accepted Answer

(d) 100 m For the same u range, R∝sin(2θ) .So, R2​R1​​=sin(2θ2​)sin(2θ1​)​ ⇒R2​=50×sin(30)sin(90)​=100 m

Question 7

Two projectiles A and B are projected with angle of projection 15° for the projectile A and 45° for the projectile B. If RA​  and RB​  be the horizontal range for the two projectiles, then:

Accepted Answer

(d) The information is insufficient to decide the relation of RA​  with RB​ .Horizontal range for the projectile, R=gu2sin(2α)​ Information of the initial velocity is not given in the question.

Question 8

A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s.(a) Find the time it takes to reach the ground .(b) Find the horizontal distance it travels before reaching the ground .(c) Find the velocity (direction and magnitude) with which it strikes the ground.

Accepted Answer

It is a case of projectile fired horizontally from a height.angle of 66° with the h=100 m, g=9.8 m/s2 (a) Time taken to reach the ground t=(2 h/g)​ =9.82×100​​=4.51sec (b) Horizontal range x=ut=20×4.5=90 m .(c) Horizontal velocity remains constant through out the motion.At A, V = 20 m/s AVy​=u+at=0+9.8×4.5=44.1 m/s Resultant velocity Vr​=(44.1)2+202​=48.42 m/sTanβ=Vx​Vy​​=2044.1​=2.205⇒β=tan−1(2.205)=66∘​ The ball strikes the ground with a velocity 48.42 m/s  at an angle 66∘  with horizontal.

Question 9

A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. (a) Find the maximum height reached.(b) Find the range of the ball. Take g = 10 m/s2.

Accepted Answer

u=40 m/s,a=g=9.8 m/s2,θ=60∘  Angle of projection.a) Maximum height h=2 gu2sin2θ​=2×10402(sin60∘)2​=60 m b) Horizontal range X=(u2sin2θ)/g=(402sin2(60∘))/10=803​ m .

Question 10

In a soccer practice session, the football is kept at the centre of the field 40 yards from the 10 ft high goalposts. A goal is attempted by kicking the football at a speed of 64 ft/s at an angle of 45° to the horizontal. Will the ball reach the goalpost?

Accepted Answer

g=9.8 m/s2,32.2ft/s2;40yd=120ft horizontal range x=120ft,u=64ft/s,θ=45∘ We know that horizontal range X=ucosθt ⇒t=ucosθx​=64cos45∘120​=2.65secy=usinθ(t)−1/2gt2=642​(2.65)1​−21​(32.2)(2.65)2​ =7.08ft  which is less than the height of goal post.In time 2.65, the ball travels horizontal distance 120ft  (40 yd) and vertical height 7.08ft  which is less than 10ft . The ball will reach the goal post.

Question 11

A popular game in Indian villages is goli which is played with small glass balls called golis. The goli of one player is situated at a distance of 2.0 m from the goli of the second player. This second player has to project his goli by keeping the thumb of the left hand at the place of his goli, holding the goli between his two middle fingers and making the throw. If the projected goli hits the goli of the first player, the second player wins. If the height from which the goli is projected is 19.6 cm from the ground and the goli is to be projected horizontally, with what speed should it be projected so that it directly hits the stationery goli without falling on the ground earlier?

Accepted Answer

The goli move like a projectile.Here h=0.196 m Horizontal distance X=2 m Acceleration g=9.8 m/s2 Time to reach the ground i.e.t=g2h​​=9.82×0.196​​=0.2sec Horizontal velocity with which it is projected be u.​∴x=ut⇒u=tx​=0.22​=10 m/s​

Question 12

In the following figure shows a 11.7 ft wide ditch with the approach roads at an angle of 15° with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch? Assume that the length of the bike is 5 ft, and it leaves the road when the front part runs out of the approach road.

Accepted Answer

Given:Width of the ditch =11.7ft Length of the bike =5ft The approach road makes an angle of 15∘(α) with the horizontal.Total horizontal range that should be covered by the biker to cross the ditch safely, R=11.7+5=16.7ft Acceleration due to gravity, a=g=9.8 m/s=32.2ft/s2We know that the horizontal range is given byR=gu2sin2α​ By putting respective values, we get:u2=sin2αRg​=sin30∘16.7×32.2​⇒u≈32ft/s Therefore, the minimum speed with which the motorbike should be moving is 32ft/s .

Question 13

A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall?

Accepted Answer

tanθ=171/228⇒θ=tan−1(171/228) The motion of projectile (i.e. the packed) is from A . Taken reference axis at  A ∴θ=−37∘  as u  is below x -axis.​u=15ft/s,g=32.2ft/s2,y=−171fty=xtanθ−2u2x2 gsec2θ​∴−171=−x(0.7536)−2(225)x2 g(1.568)​⇒0.1125x2+0.7536x−171=0x=35.78ft (can be calculated) ​  Horizontal range covered by the packet is 35.78 ft.So, the packet will fall 228 – 35.78 = 192 ft short of his friend.

Question 14

A ball is projected from a point on the floor with a speed of 15 m/s at an angle of 60° with the horizontal. Will it hit a vertical wall 5 m away from the point of projection and perpendicular to the plane of projection without hitting the floor? Will the answer differ if the wall is 22 m away?

Accepted Answer

Here u=15 m/s,θ=60∘,g=9.8 m/s2 Horizontal range X=gu2sin2θ​=9.8(15)2sin(2×60∘)​=19.88 m In first case the wall is 5 m  away from projection point, so it is in the horizontal range of projectile. So the ball will hit the wall. In second case (22 m away) wall is not within the horizontal range. So the ball would not hit the wall.

Question 15

Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed u at an angle θ with the horizontal.

Accepted Answer

Time of flight T=g2usinθ​ Average velocity = time  change in displacement ​ From the figure, it can be said AB  is horizontal. So there is no effect of vertical component of the velocity during this displacement.So because the body moves at a constant speed of ' ucosθ′  in horizontal direction. The average velocity during this displacement will be ucosθ  in the horizontal direction.

Question 16

A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally?

Accepted Answer

During the motion of bomb its horizontal velocity u remains constant and is same as that of aeroplane at every point of its path. Suppose the bomb explode i.e. reach the ground in time t. Distance travelled in horizontal direction by bomb = ut = the distance travelled by aeroplane. So bomb explode vertically below the aeroplane.Suppose the aeroplane move making angle θ  with horizontal. For both bomb and aeroplane, horizontal distance is ucosθ  t. t  is time for bomb to reach the ground.So in this case also, the bomb will explode vertically below aeroplane.

Question 17

A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s2  and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

Accepted Answer

Let the velocity of car be u  when the ball is thrown. Initial velocity of car is = Horizontal velocity of ball.Distance travelled by ball B Sb​=  ut (in horizontal direction) And by car Sc​=ut+1/2at2  where t→  time of flight of ball in air. ∴  Car has travelled extra distance Sc​−Sb​=1/2at2  Ball can be considered as a projectile having θ=90∘  ∴t=g2usinθ​=9.82×9.8​=2sec∴Sc​−Sb​=1/2at2=2 m ∴  The ball will drop 2 m  behind the boy

Question 18

A staircase contains three steps each 10 cm high and 20 cm wide (in the following figure). What should be the minimum horizontal velocity of a ball rolling of the uppermost plane so as to hit directly the lowest plane?

Accepted Answer

Given:Height of one step =10 cm Width of one step =20 cm Total height of the staircase =y=30 cm Total width of the staircase =x=40 cm To directly hit the lowest plane, the ball should just touch point E . Let point A  be the origin of reference coordinate. Let u  be the minimum speed of the ball.We have:x=40 cmy=−20 cmθ=0∘g=10 m/s2=1000 cm/s2∴y=xtanθ−g2u2x2sec2θ​⇒−20=−2u2800000​⇒u=200 cm/s=2 m/s Thus, the minimum horizontal velocity of the ball is 2 m/s .

Question 19

A man is sitting on the shore of a river. He is in the line of 1.0 m long boat and is 5.5 m away from the centre of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s, find the minimum and maximum angles of projection for successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.

Accepted Answer

When the apple just touches the end B  of the boatx=5 m,u=10 m/s,g=10 m/s2,θ=? ⇒⇒⇒θ​x=gu2sin2θ​5=10102sin2θ​⇒5=10sin2θsin2θ=1/2⇒sin30∘ or sin150∘=15∘ or 75∘​ Similarly for end C,x=6 m Then 2θ1​=sin−1(gx/u2)=sin−1(0.6)=182∘  or 71∘ So, for a successful shot, θ  may very from 15∘  to 18∘  or 71∘  to 75∘

Question 20

A ball is thrown upward at an angle of 30º with the horizontal and lands on the top edge of a building that is 20 m away. The top edge is 5 m above the throwing point. The initial speed of the ball in metre per second is (take g = 10 m/s2) :

Accepted Answer

The key problem like this is to treat the horizontal and vertical components of the motion separately and use the fact that they share the same time of flight.Horizontal component The horizontal component of the velocity is constant as it is perpendicular to g.So we can write:vcos30=t20​Where t is the time of flight∴v×0.866=t20​t=0.866v20​…(1)Vertical ComponentWe can use :s=ut+21​at2This becomes :5=vsin30t−21​gt2​∴5=v×0.5×t−21​×9.8×t25=v×0.5×t−4.9t2​We can substitute the value of t from (1) into the first part of (2) ⇒ ∴5=0.866vv×0.5×20​−4.9t2∴5=0.86610​−4.9t2​4.9t2=11.547−5=6.5474.9t2=6.547t2=4.96.547​=1.336t=1.336​=1.156 s​We can now put this value of t back into (1)⇒​0.866v20​=t∴0.866v=t20​=1.15620​∴v=1.156×0.86620​=19.97 m/s​

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