Question

A stone is dropped from a height of 80 m. At the same instant, another stone is projected vertically upwards from the ground with a speed of 40 m/s. The time after which they meet is (\(g = 10\ \text{m/s}^2\)):

• A. 2 s
• B. 4 s
• C. 6 s
• D. 8 s

Hint:

When two bodies move under gravity, equate their positions at the same time.

Answer 1

The Correct Option is A

Height of first stone: \[ y_1 = 80 - 5t^2 \] Height of second stone: \[ y_2 = 40t - 5t^2 \] Equating \(y_1 = y_2\): \[ 80 = 40t \Rightarrow t = 2\ \text{s} \]