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 $g = 10 m s^{- 2}$

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Solution

a. $V^{2} - U^{2} = 2 a s$
${(0)}^{2} - (20)^{2} = 2 \times (10) (- S)$
$400 = 20 S$
$s = 20$
Total height= $40 + 25 = 65$

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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

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(a)

How high will the ball rise? And

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

g = 10 m s^{- 2}

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

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20 m / s e c

from top of a multistory building the height of point from were the ball is thrown in

25 m

from the ground. (i) How height is the ball rise (ii) How long will it take before hitting the ground? Take

g = 10 m / 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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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 upward with a velocity of 20m/s−1 from the top of a multistory building. The height of the point from the ball is thrown is 25m from the ground.(a) How high will ball rise?(b) distance covered before the ball hits the ground?Take g=10ms−2

a. V2−U2=2as(0)2−(20)2=2×(10)(−S)400=20Ss=20Total height= 40+25=65

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. $V^{2} - U^{2} = 2 a s$
${(0)}^{2} - (20)^{2} = 2 \times (10) (- S)$
$400 = 20 S$
$s = 20$
Total height= $40 + 25 = 65$