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 orĢ) 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. V f = v i + at where v f is final velocity, v i is initial velocity, a is acceleration and t is the time.ġ) 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).

To derive this formula let us consider the figure given below which depicts a ball launched with.

But the question is how did we get this relation for the range of the projectile. So we can calculate the initial velocityĪnother of the kinematic equations of motion is: When a projectile is launched it takes a parabolic path and the range of this parabola is given by the relation. The acceleration is the acceleration of gravity (aka 'g') at -9.81 m/s 2. V f 2 = v i 2 + 2ax where v f is final velocity, v i is initial velocity, a is acceleration and x is distanceĬonsidering 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. One of the kinematic equations of motion is:

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. We see that the ball reached a height of 16.2 meters. In this case, we can consider only what happens in the y-direction to solve this problem.

Projectile formula