a basketball is thrown upwards on an arc at a 60 degree angle with the horizontal.If the velocity of the ball is 5m/s How fast must the thrower run to catch the ball after it is released

Answer 1

horizontal speed = 5 cos 60 = 2.5 m/s forever
so that is how fast the runner must run, like the end, but to make it hard:

Vi = initial speed upward = 5 sin 60 = 4.33 m/s
a = g = -9.81 m/s^2
v = Vi - 9.81 t
v = 0 at top
t = 4.33 / 9.81
2 t = time aloft = 2 * 4.33 / 9.81 = 0.883 s
so range = 2.5 * 0.883 = 2.21 meters
so runner runs 2.21 m in 0.883 s
= 2.5 m/s , Whew :)