Question

A stone is thrown upward with a velocity 50m/s. Another stone is simultaneously thrown downwards from the same location with velocity 50m/s. When the first stone is at the highest point, the relative velocity of the second stone with respect to the first one is
a) zero b) 50m/s c) 100m/s d) 150m/s

Answer 1

Answer:

Let us first consider a rough diagram to explain the given situation

IMAGE

WKt, $g\quad =\quad 9.8\frac { m }{ { s }^{ 2 } }$

For the stone A, when it reached the highest point, its velocity becomes 0 and acceleration = -g.

Hence, final velocity = initial velocity + (acceleration)\times time

v = u + at

0 = 50 + (-9.8)t

$t\quad =\quad \frac { 50 }{ 9.8 } = 5.10 secs$

for stone B,

u = 50 m/s, v = ?, a = +g = $9.8\frac { m }{ { s }^{ 2 } }$, t = 5

v = 50 + (9.8)(5.10)

v = 50 + 49.98

v = 99.98 m/s

Answer 2

Answer:100 m/s

Explanation: