Question 15
A stone is thrown vertically upward with an initial velocity of 40 ms-1. Taking g = 10 ms-2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Given,
Initial velocity (u) = 40 ms-1
Final velocity (v) = 0
g = 10 ms-2
Max height = ?
According to the third equation of motion,
v2 = u2 - 2gs
Note: [negative g as the object goes up]
Substituting we get,
0 = (40)2 - 2 x 10 x s
⇒ 1600 = 20s
⇒ s = = 80 m
Total Distance = sa + sd = 80 + 80 = 160 m
where, sa and sd are distance going up and coming down, respectively.
Displacement = 0 [as the stone comes to the same point from where it was thrown ].
Hence, total distance covered = 160 m and displacement = 0, as the first point is the same as the last point