2D Kinematics - Finding Initial Velocity

AI Thread Summary
The problem involves calculating the initial speed of a punted ball with a hang time of 3.50 seconds, kicked at a 69.0° angle. The vertical acceleration is -9.81 m/s², and the horizontal acceleration is 0 m/s². To find the initial velocity, the time to reach maximum height is determined to be half of the total hang time, which is 1.75 seconds. Using kinematic equations, the relationship between vertical and horizontal components of velocity allows for the calculation of the initial speed. The discussion concludes with the successful identification of the correct solution.
whoopie88
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Homework Statement


The "hang time" of a punt is measured to be 3.50 s. If the ball was kicked at an angle of 69.0° above the horizontal and was caught at the same level from which it was kicked, what was its initial speed? (Neglect air resistance.)

Homework Equations


All Kinematics Equations


The Attempt at a Solution


Acceleration (y) = -9.81 m/s^2
Acceleration (x) = 0 m/s^2
Time = 3.5 s
Vi,x = (Vi)cos69
Vi,y = (Vi)sin69

Now I don't think I have enough information to solve the problem, but I know the problem isn't written incorrectly. I don't know where to go with the information I have because I can't solve for anything.

Help please? Thanks in advance.
 
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You know the "hang time" therefore you can deduce the time it takes to reach the highest point. At that point there is something that happens with your Vy. If you use that with Vi,y = (Vi)sin69 and an appropriate kinematic equation you can solve the problem.
 
How can I deduce the time it takes to reach the maximum height? Is that just half of the hang time?
 
whoopie88 said:
How can I deduce the time it takes to reach the maximum height? Is that just half of the hang time?

The ball is caught at the same height from which it was kicked and the acceleration is constant. Energy is conserved. Do you see any reason for the way up to take more then the way down?
 
Oh, no I don't. Okay, thank you. I found the correct solution. Thank you!
 
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