Question

A particle is dropped from height 100 m and another particle is projected vertically up with velocity 50 m/s from the ground along the same line. Find out the height where two particle will meet? (take g = 10 $m/s^2$)

• A. 100 m
• B. 80 m
• C. 120 m
• D. 40 m

Answer 1

Answer: (B)

80 m

Relative displacement $=100m$

Relative acceleration $=0m/s^2$ because both have acceleration $g$

Relative initial velocity $=-50m/s$

Using relative motion,

$s=ut+\left(1/2\right)a{t}^2$

$100=50t+\left(1/2\right)\times 0\times t^2$

$t=2s$

For particle projected from ground,

$s=ut+\left(1/2\right)a{t}^2$

$s=50\times 2-\left(1/2\right)\times 10\times 2^2$

$=80m$