The tutor continues about projectile motion.

Back in my Oct 16 post, I began exploration of a projectile’s motion. The context is a golf ball shot at 55m/s, 27° elevation, over level ground. The post from October 16 resolves the velocity as follows:

v_x=49m/s
v_y=25m/s

Consider the next question:

Find the golf ball’s time in the air.

Solution:

v_y is the focus here. When the golf ball reaches the top of its flight, v_y=0.0m/s. Moreover, that’s at exactly half the flight time.

We find the time it takes for v_y to reach 0.0m/s:

Δv=at

which implies

t=Δv/a = 0.0-25/-9.8 = -25/-9.8=2.6s

The “up” trip takes 2.6s; returning to the ground takes the same. The golf ball’s time in the air is 2.6×2=5.2s.

I’ll be continuing about projectile motion:)

Source:

Giancoli, Douglas C. Physics. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.