# Ball thrown a distance at an angle, looking for maximum height.

1. Oct 29, 2009

### chaospoodle

1. The problem statement, all variables and given/known data
"A golfer can hit a golf ball a horizontal distance of over 300m on a good drive. What maximum height will a 301.5m drive reach if it is launched at an angle of 25 degrees to the ground?

Here's a second problem I got right:
"A quarterback throws the football to a stationary receiver who is 31.5m down the field. If the football is thrown at an angle of 40 degrees to the ground, at what initial speed must the quarterback throw the ball to reach the receiver down field? What is the ball's highest point during its flight?"

2. Relevant equations
Vi = SQRT(x(a) / sin(2*THETA))
Vy = (Vi)sin(THETA)
t = -Vi / -a
y = (Vi)sin(THETA)(t) - (1/2)at^2

a = 9.8m/s
Vi = velocity initial
Vy = velocity of Y compenent

3. The attempt at a solution
I started with Vi = SQRT(301.5m(9.8m/s) / sin(50 degrees)
which came out to be 62.11m/s, I then did Vy = (62.11m/s)sin(25 degrees)
which came out to be 26.25m/s. I then found time with the formula I posted and got 2.68s.
Then I did y = (62.11m/s)sin(25 degrees)(2.68s) - (1/2)9.8m/s(2.68s)^2
and got 35.15m. My teacher said the answer was 70.3m, which is exactly double my answer.

I was wondering why this was wrong, when I used these same equations in the second problem and got the correct answers. Both asked for the highest point of the balls during their flight. Also, both problems only gave me a distance and an angle, I assumed that would allow me to use the same equations, just plug in the different numbers.
Is my teacher wrong, or what am I doing in problem 1 wrong? Sorry if this isn't typed up correctly or too long, I was just really curious.

70.3m for the maximum height of a golf ball hit at that angle doesn't even seem right to me if we're looking at this problem literally. I'm probably wrong though.

Last edited: Oct 29, 2009
2. Oct 29, 2009

### Andrew Mason

You can verify this. The kinetic energy is the sum of the kinetic energies in the x and y directions. Energy in the x direction does not change so the potential energy at maximum height when vy =0 has to equal the initial vertical kinetic energy.

$$mgh = \frac{1}{2}mv_{yi}^2$$

so:

h = v_y^2/2g = 26.25^2/2*9.8 = 35.15m

AM

3. Oct 29, 2009

### chaospoodle

Hm, okay thanks. Going to have to explain this to my teacher now lol. He took forever thinking about the problem and coming up with a formula different from mine and didn't even really look at or think about the work I showed him.