A ball is thrown vertically upwards with a velocity of $20 m s^{- 1}$ from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.How long will it be before the ball hits the ground ? Take $g = 10 m s^{- 2}$ .

t=5s

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t=10s

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t=15s

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t=20s

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Solution

The correct option is A t=5s
Time to reach maximum height can be obtained from $v = u + a t$
$0 = 20 + (- 10) t$
$t = 2 s$
$s = u t + 0.5 a t^{2} = 20 (2) + 0.5 (- 10) (2)^{2} = 20 m$
Thus, total distance for maximum height is 45 m
$s = u t + 0.5 a t^{2}$
$45 = 0 + 0.5 (10) (t^{'})^{2}$
$t^{'} = 3 s$
Total time= 3+2= 5s

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A ball is thrown vertically upwards with a velocity of 20m/s from the top of a multi storey building.The height of the point from where the ball is thrown is 25m from the ground.

(a)How high will the ball rise

(b)How long will it be before the ball hits the ground.

Take g=10m/s

A ball is thrown vertically upwards with a velocity of 20 metre per second from the top of a multi storey building .The height of the point from where the ball is thrown is 25 m from the ground .

How high will the ball rise and how long will it be before the ball hits the ground ?

Take g = 10 m/s

I got X - X₀as 20 m . Is this the distance ball rised or is that X ?

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Q. A ball is thrown vertically upwards with a velocity of

20 {m s}^{- 1}

from the top of a multistory building. The height of the point from where the ball is thrown is

025.0 m

from the ground.

(a)

How high will the ball rise? And

(b)

how long will it be before the ball hits the ground?
Take

g = 10 m s^{- 2}

Q. A ball is thrown vertically upward with a velocity of

20 m / s^{- 1}

from the top of a multistory building. The height of the point from the ball is thrown is

25 m

from the ground.
(a) How high will ball rise?
(b) distance covered before the ball hits the ground?
Take

g = 10 m s^{- 2}

a ball is thrown vertically upwards with velocity of 25 m/s from the top of Tower of height 30m. how long it will take to travel before it hits the ground? g=10m/s^2

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A ball is thrown vertically upwards with a velocity of 20ms−1 from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.How long will it be before the ball hits the ground ? Take g=10ms−2.

The correct option is A t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at2=20(2)+0.5(−10)(2)2=20mThus, total distance for maximum height is 45 ms=ut+0.5at245=0+0.5(10)(t′)2t′=3sTotal time= 3+2= 5s

A ball is thrown vertically upwards with a velocity of $20 m s^{- 1}$ from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.How long will it be before the ball hits the ground ? Take $g = 10 m s^{- 2}$ .