A golf ball is hit with an initial velocity of 50 m/s at an angle of 45° above the horizontal. How far will the ball travel horizontally before it hits the ground?

Answer 1

Vo = (50m/s,45deg.).

Xo = hor. = 50cos45 = 35.36m/s.

Yo = ver. = 50sin45 = 35.36m/s.

t(up) = (Vf - Yo) / g,
t(up) = (0 - 35.36) / -9.8 = 3.61s.

t(dn) = t(up) = 3.61s.

T = t(up) + t(dn)=3.61 + 3.61=7.22s. =
Time in flight.

Dh = Xo * T = 35.36 * 7.22 = 255m. =
Hor. distance.

Alternate Method:

Dh = Vo^2 * sin2A / g.
A = 45 deg.
Vo = 50m/s.

Answer 2

hyt