In a penalty kick, a soccer player kicks the ball from ground level with an initial velocity of 25m/s, 20 degrees above the horizontal.
Assume no air resistance.
a) What is the maximum height, Ymax of the soccer ball?
b) What is the flight time of the soccer ball?
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The initial velocity of the ball is 25 m/s . Which could be resolved into 25 cos20 degree and 25 sin 20 degree in horizontal and vertical directions respectively.
Due to gravitational acceleration , g the ball looses its vertical component of its continuously at the constant rate of g per second. Therefore, the ball looses all its initial vertical velocity component at the maximum height due to the gravity. So,
Vertical velocity at the maximum height = 25sin 2- gt = 0 or, t = 25sin20/g . It takes (25sin20)/g seconds to attain maximum height.
The height y the ball attained in this time, t = (25sin20)/g seconds is given by:
y = (1/2) gt^2 = (1/2)g(25sin20/g)^2 =(25sin20)^2 /g = 7.4527 m
The flight time t, is given by:
y = (25sin20)t - (1/2) gt^2 = 0 , as y = 0 on the ground.
(2*25sin20 - t)t = 0. t = 0 is when the ball was on the ground before the kick.
t = (2*25*sin20)/g = 1.7432 sec
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