What is the speed of the ball when it leaves Sarah's hand?

In summary, the problem involves Julie throwing a ball to Sarah, who then throws it back to Julie. The question asks for the initial speed of the ball when it leaves Sarah's hand, with the given information of the ball leaving Sarah's hand at a height of 1.5 meters, reaching a maximum height of 15 meters, and taking 2.692 seconds to get directly over Julie's head. Using projectile motion equations and calculations, it was determined that the initial speed of the ball was 22.78 m/s. However, the initial angle of the throw was not given and should not be assumed.
  • #1
Alcubierre
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Homework Statement


The question I am struggling with is the second version of a similar problem so to make it easier for me to receive assistance, I'll post both parts:

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 20 m/s at an angle 49 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

Now for the part I am struggling with:

After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 15 m above the ground, and takes 2.692 s to get directly over Julie's head. What is the speed of the ball when it leaves Sarah's hand?

Given from second part:

[itex]t = 2.692 s[/itex]

[itex]y_{0} = 1.5 m[/itex]

[itex]y_{max} = 15 m[/itex]

[itex]v_{0,max} = 0 \frac{m}{s}[/itex]

Homework Equations



Projectile motion equations

The Attempt at a Solution



So this problem is asking for the initial speed of the ball when it leaves Sarah's hand. I tried three different approaches:

First approach: I used the range (40.337 m) found in a previous question and used the horizontal displacement equation

[itex]x = [v_{0}cos({\theta})]t[/itex]​

and solved for [itex]v_{0}[/itex] which gave me the wrong initial speed of 22.86 m/s.

Second approach: I used the vertical velocity equation

[itex]v_{y} = v_{0}sin({\theta}) - gt[/itex]​

And solved for [itex]v_{0}[/itex] which gave me 34.99 m/s, which was also wrong.

My third and last approach, was using the same steps as above, but I later (wrongly) realized that I could set it equal to 0 because the velocity is 0 at [itex]y_{max}[/itex] and used half of the total time of flight, and that gave me a initial speed of 17.49 m/s, which also turned out to be wrong.

After I finished writing this post, I realized that maybe I am wrong to assume that Sarah is throwing the ball back at a 49 degree angle to the horizontal like Julie. I have 2 more tries before I get the question wrong (it's an online system) and I would really appreciate if you guys could lead me in the right direction.

Thank you,

Rafael
 
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  • #2
Yes, you indeed made a wrong assumption on the angle.
If you take a good look at the values given, you should be able to calculate vx without much trouble.
Then you only need to calculate vy.
 
  • #3
So would I use 45 degrees, disregard the angle altogether, or find the angle?
 
  • #4
There is no angle given, so you shouldn't assume any angle. You can calculate vx and vy without using the angle. If you want to know the angle afterwards, you can calculate it from vx and vy.
 
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  • #5
Okay this is what I did:

We know that the horizontal speed remains constant, so

[itex]x = v_{0,x}t[/itex]

[itex]v_{0,x} = \frac{x}{t} = \frac{40.377 m}{2.692 s} = 14.99 \frac{m}{s}[/itex]​

Then for the vertical component,

[itex]v^2_{y} = v^2_{0,y} - 2g \Delta y[/itex]

[itex]v_{0,y} = \sqrt{2gΔy} = 17.15 \frac{m}{s}[/itex]​

So the speed is,

[itex]\left|v_{0} \right| = \sqrt{v^2_{0,x} + v^2_{0,y}} = 22.78 \frac{m}{s}[/itex]​

Is this correct?
 
  • #6
It is! I checked, thank you very much, Haye.
 

1. How do you find the initial speed of a ball?

To find the initial speed of a ball, you will need to use a formula that involves measuring the distance the ball travels and the time it takes to travel that distance. This formula is: initial speed = distance / time. You will also need to make sure your measurements are in the same units, such as meters and seconds.

2. What equipment do I need to find the initial speed of a ball?

To find the initial speed of a ball, you will need a measuring tape or ruler to measure the distance the ball travels. You will also need a stopwatch or timer to measure the time it takes for the ball to travel that distance. Additionally, a calculator may be helpful to perform the necessary calculations.

3. Can I use this method to find the initial speed of any type of ball?

Yes, the formula for finding initial speed can be used for any type of ball, as long as you have accurate measurements of the distance and time. However, it may be more difficult to measure the distance and time for smaller or faster moving balls.

4. How many trials should I do to find the initial speed of a ball?

To ensure accuracy, it is recommended to do multiple trials and take the average of the initial speeds calculated. The number of trials will depend on the precision of your measurements and the level of accuracy you are trying to achieve.

5. Is there any other method to find the initial speed of a ball?

Yes, there are other methods to find the initial speed of a ball, such as using high-speed cameras or sensors. These methods may provide more precise and accurate results, but they may also require more advanced equipment and technology.

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