# Help on projectile motion problems ?

Help on projectile motion problems plz?

Someone who is 1.6m tall throws a ball 41 deg above the horizontal at 9.4m/s.

A) How far away from the person will the ball land?
B) How long will it be in the air?
C) What is the ball's maximum height?

Formulas:
df = di + vi(t) + .5(a)(t^2)
others?

Attempt:
Velocity of the ball in the y = 9.4(sin41) = 6.167m/s
Time: 0 = 6.167 + (-9.8)(t); t = 0.63s
Max vertical distance = 0 + (6.167)(.63) + .5(9.8)(.63^2) = 5.83002

Is this right? I don't know if I did any of that right, and I don't know how to do the rest of the problem.

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## Answers and Replies

You're well on your way now. You've figured that given an initial y-velocity, it would be smart to consider when it reaches zero velocity due to the downwards acceleration from gravity, and you've got that right. You've also calculated the max height of the ball, only forgetting the height of the thrower.

Ask yourself now, how long does it take for the ball to drop to the ground from that height?

Thanks! This helped a lot. I think I'll be okay from here.... just have to be able to do this on my final exam tomorrow. *Gulp* lol

Best of luck! :)

I believe the initial height of the ball is 1.6 meters not 0

"Max vertical distance = 0 + (6.167)(.63) + .5(9.8)(.63^2) = 5.83002"

[PLAIN]http://img694.imageshack.us/img694/563/1111am.png [Broken]

"Time: 0 = 6.167 + (-9.8)(t); t = 0.63s"

the equation will only get you t1, theres still another t1 and t2 you also have to find

you can find t2 using the equation

x = xo + vot + (1/2)at2

and then using the "quadratic formula"

thus you get the time it takes to fall, and you can use that with one of the kinematic equations to get the horizontal distance traveled and how long it will be in the air.

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