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?
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