Question

A stone is dropped from the top of a tower 100m tall. At the same time another stone is thrown vertically upwards from the foot of the tower with a velocity 30m/s when and where the stone meet?​

Answer 1

Answer:

$t=3.33 s\\x_{from the ground} =54.34 m\\x_{from the top}=45.66 m$

Explanation:

Given:

STONE A

$v_{i} =0\\g=9.8 m/s^{2} \\y=100-x$

STONE B

$v_{i}=30 m/s\\g=-9.8 m/s^{2} \\y=x$

EQUATION 1:

$y=v_{i_{y} }t + \frac{1}{2}gt^{2} \\100-x=4.9t^{2}$

EQUATION 2:

$y=v_{i_{y} }t + \frac{1}{2}gt^{2} \\x=30t-4.9t^{2}$

SOLVING EQ. 1&2

$100-x+x=4.9t^{2} +30t-4.9t^{2} \\100=30t\\t=3.33 s$

$x=100-4.9(3.33)^{2}\\x=100-54.34\\x= 45.66$

The two stones will meet at distance of 54.34 m from the ground or 45.66 m from the top after 3.33 s.