Question

A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g = 10m/s, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer 1

Initial velocity u = 40 ms^-1
At maximum height, final velocity becomes zero i.e., v = 0
From equation of motion
v² = u² – 2gh
(0)² = (40)² – 2 × 10 × h
0 = 1600 – 20h
Total time taken by the ball to return to surface of the earth = 5 + 5 = 10s.
∴ h = \(\frac{1600}{20}\) = 80 m
Maximum height reached by the stone = 80 m
After reaching the maximum height, the stone will fall towards the earth and will reach the earth surface covering the same distance.
So distance covered by the stone
= 80 + 80 = 160 m
Displacement of the stone = 0
(because the stone starts from the earth surface and finally reaches the earth surface again i.e., the initial and final position of the stone are same