# Height of Ball with Velocity Vector

1. Feb 4, 2010

### dchan28

1. The problem statement, all variables and given/known data

A ball is projected horizontally from the edge of a table that is 1.02 m high, and it strikes the floor at a point 1.28 m from the base of the table. What is the initial speed of the ball?

Correct, computer gets: 2.81e+00 m/s

How high is the ball above the floor when its velocity vector makes a 43.6° angle with the horizontal?

2. Relevant equations

For the first question, I already figured out that the initial speed of the ball was 2.81 m/s and that the time it took for the ball to reach the ground 1.28 m away from the base of the table to be 0.456 seconds.

3. The attempt at a solution

For the second related question, I used the following methods to try and solve:

At the top (edge of table) before the ball begins to fall, the horizontal angle to the ball would be as follows:

tan(theta) = 1.02/1.28
theta = 38.55 degrees

The change in angle degrees from that first point to the point when the ball reaches the ground would be 90-38.55 = 51.45 degrees.

Since we want a change of only 43.6-38.55 = 5.05 degrees, I used a simple ratio as follows because acceleration due to gravity is constant (the change in degrees per time interval should stay constant):

51.45 degrees/0.456 seconds = 5.05 degrees/x
x = 0.045 s

I then plugged in this time interval into the following equation:

x(t) = x + vt + 0.5at^2
x(t) = 0 + 0 + 0.5(-9.8 m/s^2)(0.045 s)^2
x(t) = -9.81 x 10^-3 m

Finally, to get height:

1.02 m - (9.81 x 10^-3 m) = 1.01 m

However, this answer is apparently wrong. Any suggestions?

2. Feb 4, 2010

### rl.bhat

Initial velocity of the ball is 2.81 m/s in the horizontal direction. It remains constant through out.In the vertical direction velocity is zero.
As it falls vertical velocity increases.
At any instant, the resultant velocity makes an angle θ the horizontal such that
tanθ = Vv/Vh. Vh and θ ιs given. Find Vv.
Using appropriation kinematic equation find h.