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Projectile Motion Question

  • #26
If you through the projectile horizontally, its vertical component is zero. It land on the final spot without bouncing.
The time taken to fall 0.9 m is 0.429s.
Horizontal velocity to cover 2.5 m in 0.429 s is
Vo = 2.5/0.429s = 5.833 m/s.
To land on the same final spot, the other velocities with different angles of projection are such that their horizontal component must be 5.833 and time of flight must be 0.429. Because of the larger velocity it will move larger distance because of bouncing.
In the given problem the angle of projection is 45 degree. So the initial velocity will be such that vi*sin45 = 5.833. Find Vi.
Look at the picture...

vmrs41.jpg


To find the time for the ball to fall 1 meter where it bounces I use t=root(2h/g) and I get t=0.452s. To find the velocity when the ball bounces I use v=root(2gh) and I get v=4.43m/s. I now use that velocity with yf=yi+vt+1/2gt^2 to solve for t where yf=0.1m and yi=0m. This gives me t=0.881s. Adding the two times together I get 0.452s+0.881s=1.333s. Now to find the velocity I use v=d/t where d is 2.5m and get the initial velocity V=1.875m/s. That is in the horizontal direction. So assuming the ball was shot at 0 degrees instead of 45 that would be the answer. Right?
 
  • #27
rl.bhat
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No, it is not correct. With the horizontal velocity find the least velocity and least time to land the target.
 
  • #28
No, it is not correct. With the horizontal velocity find the least velocity and least time to land the target.
What is wrong with what I have? I don't understand what you mean by least velocity and least time??
 
  • #29
rl.bhat
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What I mean is you can reach the target by throwing the object horizontally or above the horizon or below the horizon. The velocity will be minimum for horizontal throw. In other two cases, the velocities should be greater because they have to cover larger distance. Larger the angle of projection larger will be the velocity. For all the cases the time will be the same.
In your solution of finding time, you have assumed that the initial vertical component of velocity is zero, which is not true.
 
  • #30
What I mean is you can reach the target by throwing the object horizontally or above the horizon or below the horizon. The velocity will be minimum for horizontal throw. In other two cases, the velocities should be greater because they have to cover larger distance. Larger the angle of projection larger will be the velocity. For all the cases the time will be the same.
In your solution of finding time, you have assumed that the initial vertical component of velocity is zero, which is not true.
Okay I understand that. Now assume that the ball is being launched along the horizontal, so the initial vertical component of velocity is zero. And also assume that the ball bounces before reaching the final landing point. Are my calculations correct for that scenario?
 
  • #31
rl.bhat
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Your formula for y is wrong. It should be
yf = yi + vt - 1/2*g*t^2.
 
  • #32
Your formula for y is wrong. It should be
yf = yi + vt - 1/2*g*t^2.
I let g=-9.8m/s^2 in my calculations.
 
  • #33
rl.bhat
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In the equation what is the value of yi?
After the bounce it should be zero. Take it zero and calculated the time.
 
  • #34
I now use that velocity with yf=yi+vt+1/2gt^2 to solve for t where yf=0.1m and yi=0m. This gives me t=0.881s.
Isn't that what I already did?
 
  • #35
rl.bhat
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Yes. I didn't notice.
0.1 = 4.43*t - 4.9*t^2
Or 4.9*t^2 - 4.43*t + 0.1 = 0
When I solved it I got different answer. Check it.
 
  • #36
Yes. I didn't notice.
0.1 = 4.43*t - 4.9*t^2
Or 4.9*t^2 - 4.43*t + 0.1 = 0
When I solved it I got different answer. Check it.
I get t=0.023s or t=0.881s and since I want the time for when the ball is coming back down I use t=0.881s I believe.
 
  • #37
rl.bhat
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OK.Then your answer is correct.
 
  • #38
Okay, thank you to everyone that helped out with this.
 
  • #39
rl.bhat
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Wait a minute. There is one flaw in your solution.
According to you the total time of flight is 1.333 s.
During that time the vertical displacement is 8.7 m!!
How is that? .
 
  • #40
That's assuming the ball is falling the entire time and not considering the bounce.
 
  • #41
rl.bhat
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You have taken two time intervals to calculate the time of flight, taking into account the bounce.
But the vertical space in only 1m!
 
  • #42
You have taken two time intervals to calculate the time of flight, taking into account the bounce.
But the vertical space in only 1m!
Yes I know, the ball falls 1m then bounces back up 1m and finally lands after traveling another 0.9m. This takes a total time of 1.333s.
 
  • #43
rl.bhat
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OK. Then it is alright.
 

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