a ball is thrown vertically upwards with a velocity of 4.9 metre per second . find the maximum height reached by the ball?
Answers
Answered by
2
v^2 = u^2 + 2as (3rd equation of motion)
v = 0 at peak position
u = 4.9 m/s
a = - 9.8 m/s^2 ( downward direction taken to be negative)
s = maximum height
0 = (4.9)^2 + 2(- 9.8)s
s = 4.9 x 4.9 /2x9.8
s = 4.9/4 = 1.225 m
Similar questions