Question

A stone is dropped from the top of a tower 96 m tall. at the same instant another stone is thrown vertically upwards from the foot of the tower with a velocity of 24 m/s. When & where will the two stone meets

Answer 1

Answer:x=78.4 m and t=4 s

Explanation:

Given

height of tower=96 m

A stone is dropped and at the same instant another stone is dropped with velocity 24 m/s

Let first stone travel x m in t s

therefore second stone travel 96-x m in t s

for first stone

$x=0+\frac{gt^2}{2}$------1

$96-x=24t-\frac{gt^2}{2}$-----2

add 1 & 2 we get

$96=24t$

$t=\frac{96}{24}=4 s$

$x=\frac{9.8\times 4^2}{2}$

$x=78.4 m$

thus at $t=4 s$ and $x=78.4 m$ from top they meet each other

Answer 2

Answer:

thanks for giving me the answer