YAPhysicsT: Solving for Vo on different elevations (solved)

[Ricemarine]

Basically, a ball was thrown on top of a roof of a building that is 20 meters tall. The time it took to land was 4 seconds. The angle when the ball landed on the roof (ending angle) was 60 degrees. What is a) the horizontal distance the ball traveled, b) the magnitude, and c) the angle the ball was initially thrown.

Now I'm confused as to go about this since the 20m building is blocking it. Since the horizontal distance was first, does that mean we're supposed to solve for that first or do we solve for the magnitude?...

I tried using the equation y = yo + (Vosin(theta))t - 1/2gt^2, getting ~ 16.88 m/s for Vo, using the 60 degrees for the ending angle.

Using that, I got 33.76 meters for the distance, but then when I got to the angle, it showed that something went wrong, since you can't do arcsin(1.45).

Help?
Thanks....

 

[Ricemarine]

The only hint given for this problem is to "reverse the motion as if on video tape"

 

[Ricemarine]

Hmm, comparing with a friend, they go 22.62meters for the horizontal distance.... Any other luck? ..

Thanks for those who looked...

 

[silverpig]

Use the hint. Do the problem backwards.

A ball is thrown off a 20 m high roof at an angle of 60 degrees. It's time of flight is 4 seconds. What was the initial velocity?

Just do the y-motion for starters to figure out the y-component (y0 = 20, yf = 0, a = -9.8, t = 4, vy0 = ?). Then use that y velocity and trig to get the x-velocity. Then use that x-velocity and 4 seconds to figure out the horizontal displacement. That's part a.

Part b is not too difficult. Find vyf (y-velocity when it lands on the ground). You also know vx (constant), so use trig here to get vi.

Part c is trivial trig if you have done part b.

Remember, everything in my version of the problem is time-reversed. When I say find v final, in your problem that would be the initial throwing velocity...

 

[blueshoe]

Yup, what silverpig said.

I worked it out and got the same answers for parts a and b.

For the last part you know the vertical component of the velocity, Vo*cos(60).
You must find the vertical component using V=Vo-.5at^2.

I ended up using arctan to find the angle. I think you'll find the complementary angle to find obtain the angle with the ground.

 

[Ricemarine]

Hmm, so double checking... With my previous stuff, I did it differently than you silverpig, but using my previous information.

a) 33.76m
b) 24.59 m/s
c) 70 degrees?

By the way, thanks for your help so far silverpig & blueshoe!

 

[blueshoe]

a) 33.72 m
b) 16.86 m/s
c) 82.48 degrees

Thats what I'm getting..hmm. For part b I assumed it meant the magnitude of the landing velocity. It's been a while since I've done problems like these so I may have done something wrong.
Here's what I did:

I reversed the situation like the hint said. I imagined a person throwing it from the top of the building.

0 = 20 + Vsin(60)*4 - .5(9.8)*4^2
V = 16.86 m/s
Where V is the magnitude of the velocity of the guy throwing it off the building (or the velocity at which it lands on the building).

To get the horizontal distance, find the horizontal component of velocity:
Vx = Vcos(60)
Vx = 16.86cos(60)
Vx = 8.43

horizontal distance = vt
horizontal distance = (16.86)(4)
= 33.72m

Thinking that Vfinal is when it hits the ground after the guy throws it from the top:
Vyfinal = Vyinitial + .5at^2
= (16.86)sin(60)-.5(9.8)(4)^2
=-63.79 m/s

You could now draw the x and y vectors of velocity to see which angle you need.
arctan(8.42/63.79)=7.5 degrees

90-7.5 = 82.48 degrees, the angle of landing with respect to the ground (or the angle of throwing from the bottom of the building with respect to the ground)

edit: if part b means the magnitude of velocity of when it is thrown from the ground it would be sqrt(8.43^2 + 63.79^2) = 64.34m/s

 

[Ricemarine]

Hmm, I'll double check with my other colleagues in the class.

Thanks guys.

Edit: Part b would be when it is thrown from the ground... I rechecked my answers, and I didn't add the 1/2 and the ^2 for t, giving me v = vo - gt.. So revised, I got 63.86 m/s and 82.4 degrees, which is close enough to your answer blueshoe. Thanks.