A ball of mass m is thrown straight upward from the top of a multi-storey building with an initial velocity of +15 m/s.
Find out the approximate maximum height reached by the ball?

23m

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15m

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11m

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Solution

The correct option is C 11m
Given : $u = 15$ m/s $a = - g = - 10 m / s^{2}$ (taking upward direction to be positive)
Final velocity of the ball at the highest point $v = 0$ m/s
Using $v^{2} - u^{2} = 2 a S$
$∴$ $0 - (15)^{2} = 2 \times (- 10) S$ $⟹ S = 11.25 \approx 11$ m

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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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A ball of mass m is thrown straight upward from the top of a multi-storey building with an initial velocity of +15 m/s.Find out the approximate maximum height reached by the ball?

The correct option is C 11mGiven : u=15 m/s a=−g=−10m/s2 (taking upward direction to be positive)Final velocity of the ball at the highest point v=0 m/sUsing v2−u2=2aS∴ 0−(15)2=2×(−10)S ⟹S=11.25≈11 m

A ball of mass m is thrown straight upward from the top of a multi-storey building with an initial velocity of +15 m/s.
Find out the approximate maximum height reached by the ball?

The correct option is C 11m
Given : $u = 15$ m/s $a = - g = - 10 m / s^{2}$ (taking upward direction to be positive)
Final velocity of the ball at the highest point $v = 0$ m/s
Using $v^{2} - u^{2} = 2 a S$
$∴$ $0 - (15)^{2} = 2 \times (- 10) S$ $⟹ S = 11.25 \approx 11$ m