Throwing a ball: Motion in one direction

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The discussion revolves around the physics of throwing a ball vertically from a building. Key calculations include determining the initial velocity, maximum height, and final velocity upon hitting the ground. The initial velocity was calculated to be 39.2 m/s, and the maximum height reached was 78.4 m above the building. The elapsed time of flight was found to be approximately 8.92 seconds, with a final velocity of 48.22 m/s upon impact. The acceleration due to gravity remains constant at -9.8 m/s² throughout the ball's motion.
  • #31


lNVlNClBLE said:
d)
-40 = 39.2t + (1/2)(-9.8)(t)^2
-40 = 39.2t + (-4.9)(t)^2
-4.9t^2 + 39.2t + 40 = 0
4.9t^2 - 39.2t - 40 = 0

= 39.2 plus/minus sqrt (1536.64 + 784)/9.8
= 39.2 plus/minus sqrt (2320.64)/9.8
= 39.2 plus/minus 48.17/9.8
t = 8.92 seconds

WHEW.

e) v = Vo + at
v = 39.2 + (-9.8)(8.92)
v = 39.2 + (-87.42)
v = -48.22
v = 48.22 m/s

f) 9.8 m/s^2

g) 9.8 m/s^2?
Looks good! (For the last two, you know that the acceleration is 9.8 m/s^2 downward. So you may want a minus sign.)
 
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  • #32


lNVlNClBLE said:
Phew. Long time since I used the quadratic. Can you check to see if that's right?
Also, I've never used it before in physics, so I'm assuming the whole negative part of the plus/minus can be ignored since there's no such thing as negative time.
Looks good. Yes, the negative time solution is not relevant for this problem.

EDIT: Completed the rest of the problem with that time. I assume f is also just 9.8 m/s^2. If so, are those accelerations -9.8, or just 9.8?
Using an "up = positive" convention, you'd need a minus sign.
 
  • #33


Doc Al said:
Looks good. Yes, the negative time solution is not relevant for this problem.


Using an "up = positive" convention, you'd need a minus sign.

Whew, thanks for all the help.

One more question. I started another problem of parabolic motion, and I only got stuck at one part. I need to find the total height of a ball thrown off the top of a building, when the height of the building is 45m, and the ball is thrown at 20m/s at a 30 degree angle.

What formula would I derive the Ymax from? I already solved the x distance displacement and the final velocity.
 
  • #34


remember that the velocity in the y direction is separate from the velocity in the x direction. If you find each component, then you can apply each of them to the relevant things that need to be found. vy for y direction and vx for x direction

so basically you can treat the distance it travels in the y direction in the same way you treated the ball in your first problem
 
  • #35


lNVlNClBLE said:
One more question. I started another problem of parabolic motion, and I only got stuck at one part. I need to find the total height of a ball thrown off the top of a building, when the height of the building is 45m, and the ball is thrown at 20m/s at a 30 degree angle.

What formula would I derive the Ymax from? I already solved the x distance displacement and the final velocity.
As SHISHKABOB says, you'll use the same formulas and methods. Just apply them to the vertical motion separately. Start by figuring out the y-component of the initial velocity.
 

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