Max height-2 balls-1 straight up-the other horizontal at an angle? yikes

[sddatt]

Homework Statement

A ball is thrown straight upward and returns
to the thrower’s hand after 4 s in the air. A
second ball is thrown at an angle of 67◦ with
the horizontal.
The acceleration of gravity is 9.8 m/s2 .
At what speed must the second ball be
thrown so that it reaches the same height as
the one thrown vertically? Answer in units of
m/s.

The Attempt at a Solution

*****I've been working on this problem for a few days. Here's what I have so far:

This is a 2-d question. First I need to find the maximum height of the first ball (thrown straight up) then I can plug that into the second ball's equations and find the Vo(x) of the second ball.
**Will the ball that's being thrown straight up have it's maximum height at 2s (half way between the total time?)
So I would have Y=0 + 0(1)+ (+)4.9(1^2).
Then Height would be +4.9? I think that gravity COULD be + because the ball is going up...but I don't know if that even makes sense.

Then I would put the Y into my second ball's equations:
1.) Y EQN: 4.9=0+Vo(y)sin(67)(t)+4.9(t^2)
2.) X EQN: x=0 + Vo(x)cos(67)(t)+ 0
3.) Vy EQN: 0=Vo(y) + -9.8(t)
4.) Vx EQN: Vx=Vo(x) + 0

**Substitute and solve for t: t=.___s
**Plug t into X EQN: x=Vo(x)cos(37)(.__s)
Now I am missing x and the magnitude.

I really don't know what to do with this problem. Any help is greatly appreciated! Thank you! :)

 

[Kurdt]

You need to find the initial vertical velocity the first ball has. That will be the same vertical velocity your second ball has.

 

[baba89]

Hello,

First of all, great job on working through the problem so far! It looks like you have the right idea in terms of finding the maximum height of the first ball and then using that in the equations for the second ball.

To clarify, the maximum height of the first ball will occur at the halfway point of its total time in the air, which is 2 seconds. So you are correct in using t=2 seconds in the equation for the first ball's height.

Next, you are also correct in using 4.9 m/s^2 as the acceleration due to gravity, and you can use a positive sign for gravity in this case since we are considering the upward motion of the ball.

Now, for the second ball, you have correctly identified the four equations you will need to solve for its initial velocity (Vo) and total time in the air (t). One thing to note is that in the third equation, you will need to use -9.8 m/s^2 as the acceleration due to gravity since the ball is moving downward at that point.

To solve for the initial velocity (Vo) of the second ball, you can use a substitution method. First, solve the fourth equation (Vx=Vo(x)+0) for Vo(x) and then substitute that into the second equation (x=0+Vo(x)cos(67)(t)+0). This will give you an equation for x in terms of t. Then, substitute the value of t that you found for the first ball (t=2 seconds) into this equation and solve for x. This will give you the horizontal distance traveled by the second ball.

Finally, to find the initial velocity (Vo) of the second ball, you can use the first equation (4.9=0+Vo(y)sin(67)(t)+4.9(t^2)). Since you know the value of t (2 seconds) and the value of x (from the previous step), you can solve for Vo(y) using this equation.

I hope this helps! Let me know if you have any further questions. Keep up the good work!