Projectile Motion or Motion in Two dimension

[Myung]

Homework Statement

A ball is thrown by Jed in the street and caught 2 seconds later by Jelo on the balcony of the house 15m away and 5.0m above the street level. What was the speed of the ball and the angle above the horizontal at which it was thrown?

Homework Equations

The Attempt at a Solution

My attempt for the solution is first get the V(Vertical)

2g(Vertical Max) = Vy^2 - Vyi^2
(Vy at top is 0 so...)
2(-9.8)(5)=-Vyi^2

and i got:
Vyi = 9.9 m/s

Then I want to get my Vxi

Range = Vxi (t)
Vxi = Range / t
Vxi = 15 / 2
Vxi = 7.5 m/s

Vi of the ball = sqrt[(7.5)^2 +(9.9)^2]
Vi = 12.42 m/s

To get angle I used the formula:

Vxi = Vi cos x
x = Cos-1 ( Vxi / Vi )
x = Cos-1 ( 7.5 / 12.42 )
x = 52.85 degrees

Am i doing it right? >.<

 

[Tomer]

You got it half right, but not quite.

If You'll be waiting on a balcony for a ball I'd thrown at you, it wouldn't necessarily arrive at velocity 0, as you suggested. The velocity would only be 0 once you stop it with your hand. But that's another force in action which has nothing to do with the problem. In short - the ball's velocity isn't necessarily 0 once it reaches the balcony.

you know that the initial vertical velocity, v_y, holds this equation:

height of the ball = y(t) = y_0 + v_y*t -0.5gt^2 (here g= 9.8)
(I hope you agree!)

Use the given data of the ball's final height (compared with the streets height) and travel time to get v_y.
The rest of your calculation looks good!

 

[Myung]

You got it half right, but not quite.

If You'll be waiting on a balcony for a ball I'd thrown at you, it wouldn't necessarily arrive at velocity 0, as you suggested. The velocity would only be 0 once you stop it with your hand. But that's another force in action which has nothing to do with the problem. In short - the ball's velocity isn't necessarily 0 once it reaches the balcony.

you know that the initial vertical velocity, v_y, holds this equation:

height of the ball = y(t) = y_0 + v_y*t -0.5gt^2 (here g= 9.8)
(I hope you agree!)

Use the given data of the ball's final height (compared with the streets height) and travel time to get v_y.
The rest of your calculation looks good!

Uhm, I'm a bit confused.. but in our class we assume that in this kind of problem the ball is always Vy = 0 once it reach the highest.

 

[Tomer]

Uhm, I'm a bit confused.. but in our class we assume that in this kind of problem the ball is always Vy = 0 once it reach the highest.

That's not accurate, and I'll give you two counter examples:

1. Who said that the balcony is the highest point on the trajectory of the ball? Imagine the ball traveling on a "rainbow" path, upwards, and then falls back again (while advancing), until it reaches the hands of the guy on the balcony. It could have gone up to 10 meters high, but then, after arriving it's peak height (and zero velocity), it falls downwards towards the balcony.
Do you understand how in this example the velocity of the ball isn't 0? Nobody said the balcony is the highest point of the ball's path.

2. Imagine a guy standing two meters above you. If you throw a ball with a large initial velocity, you can't expect it to have 0 velocity after 2 meters. However, if the guy above you catches the ball, he will inflict another force on it, which will "accelerate" the ball (with a negative sign) until it stops. That's not what they mean by the question - there's that instant moment in which the ball arrives to the guy's hands with a certain velocity, let's say, 4 m/s. Then the ball stops for other reasons (the force inflicted), but not because the ball reached it's maximum height.

I hope these two examples clarify your confusion. You can't assume the ball reached it's highest "potential" height (that means, where he could reach without other forces acting) unless it's stated in the question.

 

[rwisz]

Yes, your approach and calculations are correct. You have used the correct equations for projectile motion in two dimensions and have solved for the initial velocity and angle of the ball. Your final answer of 52.85 degrees for the angle above the horizontal is also correct. Good job!