First you have to find the vertical initial velocity which can be found by using the SOH CAH TOA rule:
Sin(25) x 13.9 = 5.87 m/s (3 S.F)
After you have the initial velocity you have to use a formula to work out the time take for the ball to travel in the air. This can be found by using the formula below:
Time = (Velocity final - Velocity initial) / Acceleration (Deceleration in this case since gravity is affecting the ball)
T = (Vf - Vi) / a
Then you plug the numbers in and you get
T = (0 - 5.87) / -9.81 (Negative since gravity is working against the ball)
= -5.87 / -9.81
= 0.60 Seconds (2 DP)
This is however not the final answer as this only works out how long it takes for the ball to stop moving. Right now we have worked out that it takes 0.60 seconds for the ball to reach the max height. For us to find out how long it takes for it to reach max height and then come back down we simply need to times the answer by 2
Time taken for ball to reach max height and fall to the ground = 0.60 x 2 = 1.20 seconds
Now that we have the time taken for the ball to fly up and come back down we can now work out the horizontal distance travelled by the ball
First we have to work out the horizontal initial velocity and this can be done by using the SOH CAH TOA rule again: (You could use a^2 + b^2 = c^2)
Cos(25) x 13.9 = 12.6 m/s (3 S.F)
Now that we have the initial horizontal velocity as well as the time taken to reach max height and come down, working out the horizontal distance should be easy. This can be done by using the formula below:
Distance = Velocity x Time
D = V x T
Now we just have to plug the numbers in
D = 12.6 x 1.2
D = 15.2 Metres
Whenever you make these calculations you always have to remember to say assuming there are no external factors acting at the end. This is because other factors such as wind and air friction would cause the results to be completely different. Hopefully this helps :)) and hopefully I wasn't wrong either.
