# How do I solve this projectile motion problem?

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1. Aug 12, 2018

### lolseeker2 • Member warned that the homework template must be used
The volleyball player serves the ball from pt.A w/ an initial Vo at angle theta to the horizontal. If the ball just cleared the net at pt. C and landed on the base at line B.

Determine the value of theta.

Determine the value of Vo.

Determine the time reach to the highest point at the reference from the horizontal surface.

The h from which the ball is thrown is 0.9 m.

Figure is in the link.

https://imgur.com/a/qDkzsbO

I tried solving Vox using Vox = 0.9sin90 by assuming the other angle from Vox to h to be 90 degrees.

I don't know how to get Voy to solve for the theta.

2. 3. Aug 12, 2018

### Delta2 We can assume that the distances AB,AC,CB as well as the height of point C from the ground, are all known to us?

4. Aug 12, 2018

### lolseeker2 The only variable that has a value is the height relative to the ground from which the ball is thrown by the player which is 0.9m

5. Aug 12, 2018

### haruspex The 0.9 (m) is a distance, so that is not going to produce a velocity. I have no idea what
means.
As Δ2 points out, you need to know the dimensions of a volleyball court. Look those up.

You should not have deleted the template. What relevant standard equations do you know for motion under constant acceleration?

6. Aug 12, 2018

### lolseeker2 Ah yes...my mistake...should have considered that. What equations should i use for solving theta?

7. Aug 12, 2018

### Delta2 Something else, when you say "it just clears the net at point C" , do you mean that the ball is at its highest point at point C?

8. Aug 12, 2018

### lolseeker2 yes

9. Aug 12, 2018

### Delta2 Something doesn't look quite right... For a volleyball court it is AC=CB right?

or point A is a bit out than the baseline at that side, while point B is at the baseline?

10. Aug 12, 2018

### lolseeker2 yes...i just did a diagram of it...I also assumed the height for the net is for men's.

11. Aug 12, 2018

### Delta2 If the ball is at its highest point at point C, and AC=CB then at point B it will be at height h from the ground (its from the symmetry of the situation, if it spends horizontal distance AC to reach the highest point, then will also spend another AC=CB distance to reach at height h which is the height it is initially launched).
It cant land exactly there, at point B, regardless of what we have for $v_0$ and $\theta$

12. Aug 12, 2018

### lolseeker2 Point B is at the baseline. Pt. A itself is where the player throws the ball with a height of 0.9 m from the ground.

13. Aug 12, 2018

### Delta2 is Point A exactly above the baseline at that side and at height 0.9m?

14. Aug 12, 2018

### lolseeker2 The ball was thrown from pt. A via player's hand at a height of 0.9m from the ground. The ball reach max height at pt. C when it cleared the net and then it dropped exactly at pt. B baseline. That's how I pictured out problem.

15. Aug 12, 2018

### lolseeker2 yes

16. Aug 12, 2018

### lolseeker2 17. Aug 12, 2018

### Delta2 This scenario isn't possible unless we account for air resistance. Are we to neglect air resistance or not?

18. Aug 12, 2018

### lolseeker2 yes

19. Aug 12, 2018

### Delta2 if we neglect air resistance, you scenario just cannot happen.

20. Aug 12, 2018

### lolseeker2 Well the teacher who gave the problem stated there is no air resistance involved.

21. Aug 12, 2018

### Delta2 @haruspex what do you think, if AC=CB and the ball is at its highest point at C, isn't at point B gonna be also at height h above the ground? (air resistance neglected)