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.)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?
Throwing a ball: Motion in one direction
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Looks good. Yes, the negative time solution is not relevant for this problem.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.
Using an "up = positive" convention, you'd need a minus sign.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?
lNVlNClBLE
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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.
SHISHKABOB
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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
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.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.
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