Projectile motion- solve for initial velocity

[mbphysics]

Homework Statement

A ball is thrown upward at an angle of 30degrees to the horizontal and lands on the top edge of a building that is 20m away. The top edge is 5.0m above the throwing point. How fast was the ball thrown?

Homework Equations

x=v₀t+(1/2)at²
v_f²=V₀²+2ax
v_x=v₀cosθ
v_y=v₀sinθ

The Attempt at a Solution

I'm stumped on this question because I feel like I should solve for time first to get initial velocity but I can't figure out how to do that without actually having the initial velocity. I solved for t using t=20/v₀cos30 and then plugged that into the v_f² equation but it comes out as a huge mess.

I know the answer is 20m/s but I don't know how to get the answer.

Thanks for any help!

 

[mfb]

Your equation for x looks odd, should that be y? And what about the other coordinate?

You have two conditions - the height and width - and two unknowns, time and initial velocity. Can you write down two equations that just depend on those (and the known angle)?

 

[TSny]

Hello and welcome to PF!

I solved for t using t=20/v₀cos30 and then plugged that into the v_f² equation but it comes out as a huge mess.

I don't see how you could substitute for t in the v_f² equation. That equation doesn't contain t. But you have the right idea. You need to substitute for t in the appropriate equation where you can use your information about the vertical height y.

[Sorry. I see mfb posted while I was constructing my post.]

 

[Chris B]

You're definitely on the right track, but v_f² doesn't involve time, does it? Basically, you've solved for the time it takes for the ball to go 20 meters with initial velocity v₀. Now, you also know that the ball has to travel 5 meters vertically in that same amount of time. Can you use that information to solve for v₀ by plugging what you have for t into an equation for the vertical motion?

 

[mbphysics]

I used t=20/v₀cos30 and plugged that into t in the equation: v_fy=v_0y+a_yt so it becomes v_fy=v_0y+a_y(20/v₀cos30) Then I found v_fy to be (v_0y²-(2*-9.8*5))^1/2 and plugged that whole thing into the vf=vo+at formula. So: (v_0y²-(2*-9.8*5))^1/2=v_0y+(-9.8)(20/v₀cos30) That way the only variable is v_0y but I don't know where to go from there

 

[mbphysics]

Is there a simpler way to solve this that I'm missing? In my practice book it is only a level 2 problem (out of 3) but it seems very complicated

 

[PeroK]

Is there a simpler way to solve this that I'm missing?

Trying to find the final vertical velocity is not necessary. Instead, think about the y displacement.

 

[Chris B]

I used t=20/v₀cos30 and plugged that into t in the equation: v_fy=v_0y+a_yt so it becomes v_fy=v_0y+a_y(20/v₀cos30) Then I found v_fy to be (v_0y²-(2*-9.8*5))^1/2 and plugged that whole thing into the vf=vo+at formula. So: (v_0y²-(2*-9.8*5))^1/2=v_0y+(-9.8)(20/v₀cos30) That way the only variable is v_0y but I don't know where to go from there

I'd say step back and think about what you're trying to solve for. You want v₀. So you have two unknowns really: time and the initial velocity. When you have two unknowns you're going to need two equations that involve those two unknowns to solve for them both. You're making things more complicated by involving a third equation, I think.

You can use the initial and final y velocity equations to solve I believe, but remember you need everything to be in terms of v₀ in order to solve for v₀. OR, you need everything in terms of v_0y. You have equations relating the two, so you can do either by subsituting.

PeroK's hint is correct though, this problem is much easier if you simply use two equations of motion for displacement, one in the x direction and one in the y direction.

This might make things a little clearer

y = v_0yt+1/2at²
x = v_0xt

Use:
v_0y = v₀ sinθ
and
v_0x = v_o cosθ

To put everything in your displacement equations in terms of v₀ and the rest is algebra (which might be a little tricky, but if you write down all your steps shouldn't be too bad).