A ball of mass m is thrown from the ground at the speed u=10ms^-1 at an angle of 30 degrees. Find the max height, the total flight time and the max distance it travels?Assume g=10ms^-1 and there is no air friction
First let’s split the problem into 2 parts: Vertical(y direction) and Horizontal(x direction.Vertical: u_y = 10 sin (30) = 5ms^-1
As for the ball to stop at max height and the fall down, all the kinetic energy has to be converted into potential energy: KE=PE1/2 mu_y^2 = mgh_max so h_max= v^2/2g = 5/4 m
The time of flight can be found by considering the time required to reach the max height and double it as we know that the time for the ball to reach max_height will the same as the time for the ball to fall from the max_height to the ground.
0 = u_y - gt, the minus comes from the fact that the acceleration acts in the opposite way of the vertical velocity
t=u_y/g = 0.5s so T_total = 2*t = 1s
Horizontal: u_x = 10 cos(30) = 5 * 3^1/2 ms^-1
Max distance can be found once we had found the total flight time as we know that the ball must travel 1s in the x direction having a velocity of u_x
So D_max= u_x * T_total = 5 * 3^1/2 m
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