Help with homework question

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In summary, the conversation discusses finding the magnitude of the initial velocity of a tennis ball thrown above the horizontal, given that it takes 0.50 seconds for the ball to reach the top of its trajectory. The experts suggests using the equation v = a*t to calculate the vertical component of velocity, which is equal to 4.9 m/s. They also mention using the equation R = (Vo*cosθ)*t to find the range, which in this case is 8.5m. Finally, they suggest using the equation Vy/Vx = tanθ to find the angle, and mention that it could have been simpler to use the equation Vo = 2*4.9 if the angle was known. The conversation ends with
  • #1
nelly42688
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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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  • #2
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.
 
  • #3
so basically magnitude would be 4.9m/s since your dividing gravity by .5?
 
  • #4
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:
  • #5
nelly42688 said:
so basically magnitude would be 4.9m/s since your dividing gravity by .5?

This is correct.
 
  • #6
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?
 
  • #7
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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