a ball is launched at an angle to the ground with an initial speed of 65m/s. at it's maximum height it has a speed of 26m/s. determine it's maximum height.

Answer 1

At max height all velocity is horizontal

u = 26
but u = 65 cos Theta
so Theta, the angle shot above horizontal = 66.4 deg
so
Vi = initial up speed = 65 sin Theta = 59.6
at top vertical speed = 0
so
0 = 59.6 - 9.8 t where t is rise time
t = 6.08 seconds going up
H = 0 + 59.6 t - (9.8/2) t^2
h = 59.6(6.08) - 4.9 (6.08)^2
h = 362 - 181
= 181 meters

Answer 2

A ball Is launched at 5.5 m/s at 76 above the horizontal it starts and lands at the same distance from the ground what are the maximum height above its launch level and the flight time of the ball