Question

A ball was thrown with a velocity of 150 ft/ s at angle of 45 degrees above the horizontal.

a)Find the range and the time of flight.
b)What is the maximum height reached?

Answer 1

a) Let the angle be A and the range R.
g = 32.2 ft^2/s.
T = time of flight

R = (V^2/(2g)) sin(2A)
T = 2*V*sinA/g
Note that 2A = 90 degrees so
sin(2A) = 1. Thus R = V^2/(2g) @ A=45
b) Max. height H = (V sinA)^2/(2g)

Answer 2

Thanks drwls...