How Do You Calculate the Initial Velocity in Horizontal Projectile Motion?

[Physics_Newbie]

Well, this is my first of most likely many posts. Here goes the homework problem:

An Engineering student wants to throw a ball out of a third story window (10 m off the ground) onto a target on the ground placed 8.0m away from the building. (a) If the student throws the ball horizontally, with what velocity must the ball be throw? (b) What must the velocity of the ball be if it's thrown up at an elevation angle of 29degrees? (c) What is the ball's time of flight in case of (b)?

I understand how to do both parts (b) & (c) but part (a) is giving me a bit of trouble.

The only way that I've figured out how to do part (a) is to make a parabola out of the figure then figure out the horizontal component at the max height (of 10m). This seems to be a roundabout way of completing the problem.

So, if anyone has a suggestion please share.

Thank You

 

[Doc Al]

Welcome to PF!

I understand how to do both parts (b) & (c) but part (a) is giving me a bit of trouble.

Problem a is a simpler version of problem b! So I don't understand how you could do b, but not a. Show what you've done.

One way to do projectile motion problems is to treat the x and y motions separately. What are the relevant kinematic equations?

 

[Physics_Newbie]

Well, what's throwing me off is that theta=0 degrees @ T=0

So, when I plug in that value into the range equation it's spitting back 8(-9.8m/s^2)= (Vr)^2 x sine(2*0). This is obviously giving me the wrong answer since 8 is not equal to zero.

I'll print out the instruction sheet for the LaTeX script so I can start typing these out more clearly for everyone involved.

 

[Doc Al]

That range equation only applies on level ground, not when tossing a ball out the window. Forget it.

Set up the kinematic equations for vertical motion (uniform acceleration) and horizontal motion (constant speed). Hint: when you toss the ball horizontally, its vertical component of velocity is zero.

 

[Physics_Newbie]

Ah, that makes sense.

Thanks for your patience & help.