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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