How Do You Calculate the Initial Velocity of a Tennis Ball Thrown Upward?

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To calculate the initial velocity of a tennis ball thrown upward, the vertical component can be determined using the equation v = a*t, where acceleration (a) is -9.8 m/s² and time (t) is 0.5 s, resulting in a vertical velocity of 4.9 m/s. The total flight time is 1 second, which helps clarify the confusion about the initial velocity magnitude. The range can be calculated using R = (VoCOS(θ))T, leading to a range of 8.5 m and a horizontal velocity (Vx) of 8.5 m/s. The initial velocity (Vo) can then be found using the formula (Vx² + Vy²)¹/², and the angle θ can be derived from the relationship Vy/Vx = tanθ.
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1. A tennis ball is thrown from ground level with velocity directed above the horizontal. If it takes the ball 0.50 s to reach the top of its trajectory, what is the magnitude of the initial velocity? You should draw a diagram to show this problem in your explanations.


i no that Ax=0 Ay=-9.8 i have all the equations just I don't know which to use, once i figure taht out i should be fine any help would be great
 
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Welcome to PF.

If you know how long it took for acceleration to reduce the velocity to 0 at the top of the trajectory, then you know the vertical component of velocity.

v = a*t

You know a and t.

But without further info about what happens in the x direction or the angle, that's about all you can tell is the vertical component.
 
so basically magnitude would be 4.9m/s since your dividing gravity by .5?
 
or would it be 9.8 since the total flight path is 1 second... this is where i get confused

Now for finding the ragne i know the R= (VoCOS(30))T which in this case would be 8.5m

I know i didnt explain well how i got these figures but its hard to type it all out on a keyboard. any help would be great and thank you in advance
 
Last edited:
nelly42688 said:
so basically magnitude would be 4.9m/s since your dividing gravity by .5?

This is correct.
 
nelly42688 said:
or would it be 9.8 since the total flight path is 1 second... this is where i get confused

Now for finding the ragne i know the R= (VoCOS(30))T which in this case would be 8.5m

I know i didnt explain well how i got these figures but its hard to type it all out on a keyboard. any help would be great and thank you in advance

Now that you have revealed that the range is 8.5m then you can use the total time - which is twice the time to max height to determine the Vx. In this case since time is 1 sec, that makes Vx 8.5 m/s.

So there you have it.

The initial V is (Vx2 + Vy2)1/2

And if you didn't know θ, the angle is given by Vy/Vx = tanθ

Since it turns out you knew θ, you could have simply said 4.9 = Vo*sin30 or Vo = 2*4.9

Maybe next time post the whole problem?
 
ah sorry i thought i put θ in there, sorry about that but thank you so much for your help, you answered a lot of questions for me that i just didnt understand.
 
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