Projectile Motion - Baseball

  • Thread starter 1irishman
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  • #1
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Homework Statement


Note: zero air resistance for this problem.
A home run is hit and the ball lands in the seats at a height of 7.5m above the point at which it was hit. The ball had a velocity of 36m/s at 28deg above the horizontal as it left the bat.
a) What max height did ball reach?

b) How long did ball spend in the air?

c) How far did it travel horizontally?

d) What was the smallest speed the ball had during its flight?


Homework Equations


Vf^2 = Vi^2 + 2ad
d= Vit + 1/2at^2


The Attempt at a Solution


I got these answers
a)14.6m
b)2.54s
c)32m horizontal
d) at the top of its trajectory speed was 0m/s

Could anyone please tell me if these numbers are right? And whatever ones are wrong please give me some hints to solve? Thank you.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Show your calculation for (b)
Your answer for (d) is wrong. Particle does not stop at the top of the trajectory.
 
  • #3
Show your calculation for (b)
Your answer for (d) is wrong. Particle does not stop at the top of the trajectory.
I think his answer for (d) is correct. The y-axis component of velocity does (momentarily) become zero. The only the x-axis component of velocity does not change. I believe the problem could have been more specific, but in one sense, he is correct.
 
  • #4
rl.bhat
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I think his answer for (d) is correct. The y-axis component of velocity does (momentarily) become zero. The only the x-axis component of velocity does not change. I believe the problem could have been more specific, but in one sense, he is correct.
The answer is correct but the reasoning is wrong.
 
  • #5
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My calculation for (b):
Given d=vit + 1/2at^2
-14.6=16.9t + 1/2(-9.8)t^2
-14.6=16.9t - 4.9t^2
-14.6=t(16.9 - 4.9t)
-14.6 - 16.9= -4.9t^2
-31.5/-4.9= -4.9t^2/-4.9
t^2=sqrt of 6.43
t = 2.54s

(d) At the top of the baseballs trajectory velocity was momentarilly 0m/s. No? Please help me reason this so i can conceptualize it better? Am I to assume the answers for a and c are correct since you only pointed out b and d? thank you for your help.
 
  • #6
rl.bhat
Homework Helper
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At what height the ball lands?
It is above the point at which it was hit..
So d is positive.
d) Only vertical component of velocity is momentarily zero. But horizontal component of velocity is present there. And that is the minimum speed.
I have not checked c.
Please note! Instead of giving the answers, if you show your calculations, we can point out your mistakes. Other wise we have spend a lot of time to solve the problem and check the answer.
 
Last edited:
  • #7
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My calculation for (b):
Given d=vit + 1/2at^2
-14.6=16.9t + 1/2(-9.8)t^2
-14.6=16.9t - 4.9t^2
-14.6=t(16.9 - 4.9t)
-14.6 - 16.9= -4.9t^2
-31.5/-4.9= -4.9t^2/-4.9
t^2=sqrt of 6.43
t = 2.54s

(d) At the top of the baseballs trajectory velocity was momentarilly 0m/s. No? Please help me reason this so i can conceptualize it better? Am I to assume the answers for a and c are correct since you only pointed out b and d? thank you for your help.
Where did you get -14.6 for vertical displacement?
 
  • #8
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sorry...it should have been up the y-axis so +14.6m for vertical displacement. Is that right?
 
  • #9
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sorry...it should have been up the y-axis so +14.6m for vertical displacement. Is that right?
Well, if you're solving for the total time the ball's been up in the air, then the grand total vertical displacement of the ball is +7.5 meters.
 
  • #10
rl.bhat
Homework Helper
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sorry...it should have been up the y-axis so +14.6m for vertical displacement. Is that right?
No. Read the problem again. See my previous post.
 
  • #11
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Well, if you're solving for the total time the ball's been up in the air, then the grand total vertical displacement of the ball is +7.5 meters.
But the max height the ball reaches is 14.6m right?
 
  • #12
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But the max height the ball reaches is 14.6m right?
Yes, it is, however, plug in 7.5 for d into the equation d = V1*t + 0.5*a*t^2 and solve for overall time the ball was in the air to correctly solve for the horizontal displacement.
 
  • #13
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oh yes okay i see i think:
i got t=2.41s
and horizontal distance is 58.3m
is that right?
 
  • #14
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Actually, I believe you've made a mistake somewhere, so here's my solution:

d = V1*t + 0.5*a*t^2

7.5 = -4.9*t^2 + 36*sin28 (to make it a beautiful quadratic equation):
0 = -4.9*t^2 + 36*sin28 - 7.5
Plug all that into a quadratic formula or a calculator, and you get two values:

t = 0.523 s
t = 2.93 s

The 0.523 second time is the time it takes the ball to reach the height of 7.5 m as it's still going up, and 2.93 s is the time it takes it to land. 2.93 is the time we want here.
 
  • #15
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Solve for horizontal displacement using the horizontal component of the initial velocity and the overall time of flight.

2.93s x 36*cos28 m/s = 93 m.
 
  • #16
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Oh I think i used the x-component instead of the y-component here...i'll show you so you can hopefull tell me what values i ignored? Here goes:
d= vi t +.5at^2
7.5=36t + .5(-9.8)t^2
7.5=36t-4.9t^2
then with the rest of algebra solving for t i got t = 2.41s
then for horizontal i went like this:
d=36(2.41) + .5(-9.8)(2.41)^2
d= 58.3m
 
  • #17
29
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Oh I think i used the x-component instead of the y-component here...i'll show you so you can hopefull tell me what values i ignored? Here goes:
d= vi t +.5at^2
7.5=36t + .5(-9.8)t^2
7.5=36t-4.9t^2
then with the rest of algebra solving for t i got t = 2.41s
then for horizontal i went like this:
d=36(2.41) + .5(-9.8)(2.41)^2
d= 58.3m
Remember, the vertical component does not equal 36 m/s. 36 m/s is the velocity at 28 degrees to the horizontal, so its component must be 36*sin28.
 
  • #18
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How do you plug the value of 36*sin28 into the quadratic formula?!
 
  • #19
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How do you plug the value of 36*sin28 into the quadratic formula?!
-4.9*t^2 + 36sin28*t - 7.5 = 0

b = 36sin28

t = -36sin28+-SQRT[(36sin28)^2 - 4*(-4.9)*(-7.5)]
.......................................2*(-4.9)

Best done on a calculator.
 

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