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?

Answer 1

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,

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

⇒$h=v\times 0.204v-12\cdot 9.8\times (0.204v{)}^2$

⇒$h=0.204v^2-4.9\times 0.041v^2$

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

$h=0.004v^{2}m$