Ball thrown to cliff (projectile problem)

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Homework Statement



A man standing 31 m from the base of a vertical cliff throws a ball with a speed of 38 m/s aimed directly at a point 5 m above the base of the cliff.

Here is a link to the diagram....

https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-04-2D-Motion/ball_to_cliff/8.gif [Broken]

How long does it take the ball to reach the cliff?

Homework Equations



First off i tried to find the time to get to the max height:
v = v(initial)+at
0=38+(-9.8)t t=3.87755

Then I found the max height:
x(final) = x(initial) + v(initial)t + 0.5at^2
x(final) = 0 + 38(3.877) + 0.5(-9.8)(3.877^2) x(final) = 73.67

This really does not look right to me though. If it is then would I have to find the time it drops from the max height to 5m? Is there anyway of finding the y component?
 
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Answers and Replies

  • #2
kuruman
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You need to break the initial velocity into horizontal and vertical components before attempting a solution.
 
  • #3
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I was thinking that. But my next question would be how? The angle is not given. Would I use the distance of 31m and the height of 5m to find it?
 
  • #4
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I took the arctan of (5/31) and got 9.15 degrees. Im thinking that is the and at which the ball was thrown. Does that sound right? I think after I figure this part out I would be able to solve the problem.

Thanks!
 
  • #5
kuruman
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Yes arctan(5/31) will give you the angle of projection. Note that you don't really need to find the actual angle. All you need is the sine = opp/hyp and the cosine = adj/hyp. Use the Pythagorean Theorem to find hyp.
 
  • #6
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Ok I figured out the time, which turned out to be 0.82s. And it was simply the velocity over the distance.


i figured wow fast is the ball moving when it reaches the cliff by using the pythagorean theorem, and it turned out to be 37.3m/s.


Here is where I am having trouble...

At what time does the ball reach its largest vertical height?

I am sure that I have to find the initial vertical velocity:

31 = 38cos(x)(0.82) where x is the angle x = 5.8 degrees


vy0 = 38sin(x) vy0 calculated to be 3.84

vf = v0 + at
0 = 3.84 + (-9.8)t = 0.39s

time to max height i figured to be 0.39s

This is the incorrect answer, what did I do wrong? Any suggestions?

Thanks!
 
  • #7
kuruman
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I don't know how you calculated the angle that you call x, but it is incorrect. If I were doing this problem, I would not bother with the angle. For example, if I needed the sine of the angle, I would use

[tex]sin\theta=\frac{opp}{hyp}=\frac{5}{\sqrt{5^{2}+31^{2}}}[/tex]

Less calculation, less likelihood for error, you know what I mean?
 
  • #8
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That is understood. Perhaps I am over complicating this one. It seemed so easy to find the height but I end up with the wrong answer. Thanks for the input!
 
  • #9
kuruman
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Your method for finding the height by first finding the time it takes to reach that height is correct and should work. However, you can do it in one step if you use the kinematic equation that does not involve time,

2aΔy=vy2-v0y2

You know or can calculate everything except for the unknown vertical displacement.
 
  • #10
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Thanks for the help! It worked and I managed to find everything else that I was asked to find.
 

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