A ball is thrown vertically upwards from the top of a tower with a velocity of ${10 m s}^{- 1}$ . The ball reaches the ground with the velocity ${30 m s}^{- 1}$ . What is the height of the tower? $(T a k e g = {10 m s}^{- 2})$

40 m

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20 m

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60 m

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80 m

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Solution

The correct option is A 40 m
Taking downward direction to be positive.
Initial velocity $u = - 10 m / s$
Final velocity $v = 30 m / s$
Let the height of the tower be $H$ .
Acceleration $a = g = 10 m / s^{2}$
Using $v^{2} - u^{2} = 2 a H$
$∴$ $30^{2} - (- 10)^{2} = 2 \times 10 \times H$
$⟹ H = 40 m$

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A ball is thrown vertically upwards from the top of a tower with a velocity of 10ms−1. The ball reaches the ground with the velocity 30ms−1. What is the height of the tower? (Takeg=10ms−2)

The correct option is A 40 mTaking downward direction to be positive.Initial velocity u=−10 m/sFinal velocity v=30 m/sLet the height of the tower be H.Acceleration a=g=10 m/s2Using v2−u2=2aH∴ 302−(−10)2=2×10 ×H⟹ H=40 m

A ball is thrown vertically upwards from the top of a tower with a velocity of ${10 m s}^{- 1}$ . The ball reaches the ground with the velocity ${30 m s}^{- 1}$ . What is the height of the tower? $(T a k e g = {10 m s}^{- 2})$