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Second Equation Of Motion And Third Equation Of Motion

An object is projected vertically upwards with a velocity 100

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Hard

Solving time: 4 mins

An object is projected vertically upwards with a velocity $100 m s^{- 1}$ . What will be the ratio of distance to displacement of the object at the end of $15 s$ ? (Given, $g = 10 m s^{- 2}$ )

3 : 5

5 : 3

13 : 15

15 : 13

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An object is projected vertically upwards with a velocity

100 m s^{- 1}

. What will be the ratio of distance to displacement of the object at the end of

15 s

? (Given,

g = 10 m s^{- 2}

)

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

-

-

Displacement,

s = u t - \frac{1}{2} g t^{2}

= 100 \times 15 - \frac{1}{2} \times 10 \times 15 \times 15

= 1500 - 1125

= 375 m

Distance,

s = AB + BC

H_{m a x} = AB = \frac{v ^{2} - u ^{2}}{- 2 g} = \frac{0 ^{2} - 10 0 ^{2}}{- 2 \times 10} = 500 m

Use,

v = u - gt \Rightarrow t = \frac{u}{g}

= \frac{100}{10} = 10 s

BC = \frac{1}{2} \times 10 \times 5 \times 5 = 125 m

\Rightarrow s = AB + BC = 500 + 125

= 625 m

\frac{s}{s} = \frac{625}{375} = \frac{5}{3}

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Question Text	An object is projected vertically upwards with a velocity $100 m s^{- 1}$ . What will be the ratio of distance to displacement of the object at the end of $15 s$ ? (Given, $g = 10 m s^{- 2}$ )
Topic	Motion
Subject	Science
Class	Class 9
Answer Type	Text solution:1
Upvotes	139

An object is projected vertically upwards with a velocity $100 m s^{- 1}$ . What will be the ratio of distance to displacement of the object at the end of $15 s$ ? (Given, $g = 10 m s^{- 2}$ )

Text solutionVerified

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An object is projected vertically upwards with a velocity 100 m s−1. What will be the ratio of distance to displacement of the object at the end of 15 s ? (Given, g=10 m s−2 )

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An object is projected vertically upwards with a velocity $100 m s^{- 1}$ . What will be the ratio of distance to displacement of the object at the end of $15 s$ ? (Given, $g = 10 m s^{- 2}$ )