Question

A wooden block of mass $10gm$ is dropped from the top of a tower $100m$ high. Simultaneously, a bullet of mass $10gm$ is fired from the foot of the tower vertically upwards with a velocity of $100m{s}^{-1}$ . If the bullet is embedded in it, how high will it rise above the tower before it starts falling? $\left(g=10m{s}^{-2}\right)$

Answer 1

Answer: 75

Block and bullet will collide after $1s$ at height $95m$ from ground.
$Velocityofblockbeforecollision=gt=10m{s}^{-1}$
$Velocityofbulletbeforecollision=u-gt=90m{s}^{-1}$
from conservation of linear momentum,
$m\left(-10\right)+m\left(90\right)=2mv\prime \Rightarrow v\prime =40m{s}^{-1}$


Thus height raised by the combined mass
$h=\frac{v{\prime }^2}{2g}=\frac{40\times 40}{2\times 10}=80m$
$heightfromtower=75m$