Formula for total air time with projectile motion?

Question

"A golf ball is hit, and reaches a maximum height of 16.2 meters over the level fairway while travelling 180 meters horizontally... The ball's total time in the air was?" If this is all the information I am given, what formulas can I use to calculate total time? My instructor's notes were confusing and difficult to read. Thank you!

Answer 1

In this case, we can consider only what happens in the y-direction to solve this problem. We see that the ball reached a height of 16.2 meters. The time in the air will be the same as if the ball was hit straight up and then came straight down so we won't even need to worry about the 180 meters.

One of the kinematic equations of motion is:

v_f² = v_i² + 2ax where v_f is final velocity, v_i is initial velocity, a is acceleration and x is distance

Considering just half of the flight of the ball, the distance (x) then is the height of 16.2 meters and the final velocity will be zero (the ball slows until it reaches its highest point then begins going back at a negative velocity. The acceleration is the acceleration of gravity (aka "g") at -9.81 m/s². So we can calculate the initial velocity

0 = v_i² + 2(-9.81)(16.2)

v_i² = 317.844

v_i = 17.828 m/s

Another of the kinematic equations of motion is:

v_f = v_i + at where v_f is final velocity, v_i is initial velocity, a is acceleration and t is the time.

We can either:

1) Consider half of the path of the ball and calculate time by then doubling the answer (half the time the ball is going up and the other half the time the ball is coming down). In this case, v_f = 0 so we get: 0 = 17.828 + -9.81t then t = 1.817 then double it to get a total air time of 3.63 s or

2) We can consider the enteire flight of the ball in which case the ball's initial velocity is 17.828 and the ball's final velocity is -17.828 (the ball hits the ground going the same speed as it left but in the opposite direction) and then the time t will be the total air time. So 17.828 = 17.828 + -9.81t or 9.81t = 35.656 so t = 3.63 s

Hope that helps.

Answer 2

For this problem, it doesn't matter how far it traveled horizontally. So long as it reached a height of 16.2 meters and took just as long to get back down, it could have been launched straight up or a mile and it still would have the same air time. Another thing to keep in mind is that, due to acceleration from gravity being constant, the time it takes to get from the ground to the peak is the same as the time it takes from getting from the peak back to the ground. Basically, just double the amount of time it takes for the ball to free fall from 16.2 meters.

X = v₀ * (t/2) + 1/2 * a * (t/2)²

X = 16.2 meters

v₀ = 0

a = 9.8m/s²

t/2 = the time to fall from the maximum height, or half the total air time.