horizontal
U = 20 cos 25 for the whole trip
so
range = 20 cos 25 * time in air
vertical
initial speed up = Vi = 20 sin 25
v = Vi - g t where t is time up
so
t = Vi/g = (20/9.8) sin 25
time down = t = time up
so
time in air = 2 t
so
range = 2 t * 20 cos 25
for max height, height at t
h = Vi t - 4.9 t^2
A ball is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. i) calculate maximum height reached by the object ii) Calculate the total flight time of the object iii) Calculate the horizontal range (maximum x above ground) of the object
1 answer