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Standard XI

Physics

Applications of equations of motion

A ball is thr...

Question

A ball is thrown vertically upwards with speed $10 {ms}^{- 1}$ from the top of a tower. It hits the ground after some time with a speed of $50 {ms}^{- 1}$ . The height of tower is $(g = 10 {ms}^{- 2})$

240 m

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120 m

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60 m

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80 m

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Solution

The correct option is B $120 m$
$v^{2} = u^{2} + 2 g h$
$h = \frac{v^{2} - u^{2}}{2 g} = \frac{50^{2} - 10^{2}}{2 \times 10} = 120 m$

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A ball is thrown vertically upwards with speed 10 ms−1 from the top of a tower. It hits the ground after some time with a speed of 50 ms−1. The height of tower is (g=10 ms−2)

The correct option is B 120 mv2=u2+2gh h=v2−u22g=502−1022×10=120 m

A ball is thrown vertically upwards with speed $10 {ms}^{- 1}$ from the top of a tower. It hits the ground after some time with a speed of $50 {ms}^{- 1}$ . The height of tower is $(g = 10 {ms}^{- 2})$

The correct option is B $120 m$
$v^{2} = u^{2} + 2 g h$
$h = \frac{v^{2} - u^{2}}{2 g} = \frac{50^{2} - 10^{2}}{2 \times 10} = 120 m$