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Question

A ball is thrown vertically downwards with velocity V from the top of a tower and it reaches the ground with speed 3V.what is the height of the tower?


Solution

u→The initial upward velocity of the ball$$=v$$

v→reaches the ground in speed$$=3v$$

g→Acceleration due to gravity$$= 9.8m/s^2$$

Applying equation of motion

$$v=u+at$$

$$3v=v+9.8\times t$$

$$3v-v=9.8t$$

$$2v=9.8t$$

$$\frac{2v}{9.8}=t$$

$$0.204v=t$$

 Now,

$$ h=u\times t+12g\times t^2$$

$$h=v\times 0.204v−12⋅9.8\times(0.204v)^2$$

$$h=0.204v^2-4.9\times0.041v^2$$

$$h=v^2(0.204-0.200)$$

$$h=0.004v^2m$$


Physics

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