A ball is kicked with an initial velocity of 20 m/s at an angle of 30^{o} above the horizontal. Calculate the following:
initial speed = 20m/s
initial angle = 30 degree
a) using conservation of enrgy tells us that the speed with which the ball hits the ground is same as that with which it was thrown because there is no change in the potential energy.
Thus , the ball hits the ground with a speed = 20 m/s
b) using the formula for the time of flight of a projectile = 2U*Sin(theta)/g
= 2*20*Sin(30)/9.8
= 2.04 seconds
c) using the formula for maximum height of a projectile = [Usin(theta)]^2/2g
H_max = [20*sin(30)]^2/(2*9.8)
H_max = 5.1 m
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