# Kinematics projectile question

1. Sep 6, 2007

### fizzzzzzzzzzzy

At what projection angle will the range of a projectile equal its maximum height?

3. The attempt at a solution
alright so i figure that i can just solve the equation be figuring out at what time the projectile will be at its actual height:
d = v*sin(x)t - 9.8/2 * t^2
d' = v*sin(x) - (9.8)*t
0 = v*sin(x) - 9.8t
9.8t = v*sin(x)
t = (v*sin(x))/9.8
finding out how far horizontally the projectile will go in that time:
d = v*cos(x)*(v*sin(x))/9.8
make the 2 distances equal:
v*cos(x)*(v*sin(x))/9.8 = v*sin(x)*(v*sin(x))/9.8 - 9.8/2 * (v^2)*(sin(x))^2/(9.8)^2
simplify:
cos(x) = sin(x)/2
sqrt(1-sin(x)^2) = sin(x)/2
1- sin(x)^2 = (sin(x)^2)/4
1= 5/4 * sin(x)^2
4/5 = sin(x)^2
sqrt(4/5) = sin(x)
arcsin(sqrt(4/5)) = x
and that should give me the right velocity right?
that gives me 63.4 degrees but the answer in the book is 76 degrees

whats wrong?

2. Sep 6, 2007

### qspeechc

I haven't the foggiest idea what you were doing in your calculations, perhaps add explanations next time.

Here's what I would do.
Find an expression for the maximum vertical height for a given angle x. We know how to get this from the equations of motion, particularly when the velocity in the vertical direction is zero.
Now find the expression for the horizontal distance travelled for this angle x. Equate this distance with the vertical distance. Solve for the angle x.

3. Sep 6, 2007

### drpizza

I'll look for it in a second, but there's an equation that can be derived for the range of a projectile... you essentially tried to derive that equation. However, you're off by a factor of 2.

Before I look, I'm almost going to assume that you forgot that 1969 #1 hit by Blood, Sweat, and Tears "Spinning Wheel"

"What goes up, must come down..."

Last edited: Sep 6, 2007
4. Sep 6, 2007

### drpizza

There you go! Hopefully that helps.

edit: and yes, it rounds off to 76 degrees.

Last edited: Sep 6, 2007
5. Sep 6, 2007

### fizzzzzzzzzzzy

thanks drpizza youve been the best help so far, but theres one part of what you said that i dont get. You said that the first equation was alright as equaling d, but after i plugged in t in that equation you said that the right side had to be fixed. I'm not quite sure what you mean when you explain why its not right cause v is supposed to represent the overall initial velocity.

thanks

6. Sep 6, 2007

### drpizza

I think there are two parts you may have misunderstood... the equation for d *is* correct for the horizontal distance traveled when the object is at it's highest point. But, the object isn't done traveling yet! It's going to travel just as far horizontally while it's falling back to earth.

Now, off to the other side:

I'll give you a similar problem to think about, with actual numbers:
Initial velocity is 80 meters per second, fired at an angle of 30 degrees.
when you find the vertical component of velocity, it's 40 meters per second.

Now, you can plug those numbers in (I'm using 10 m/s^2 for gravity if you don't mind) you'll find that the time to the maximum height is 4 seconds.

4 seconds to maximum height, object launched is initially traveling 40m/s in the vertical direction. You essentially multiplied those two quantities together to get a vertical height of 160 meters. But, that 40 m/s that you started with is going to be decreasing at a rate of 10m/s every second. At the maximum altitude, it'll be going 0m/s in the vertical direction. So really, it's maximum altitude in this case would be 20m/s * 4s = 80 meters.

7. Sep 6, 2007

### learningphysics

Right side is fine. Left side needs to be multiplied by 2.

8. Sep 6, 2007

### learningphysics

drpizza, fizzzy used d = v1*t + (1/2)at^2.... not d = v1*t

9. Sep 7, 2007

### drpizza

Ahhh, shame on me for going through that so quickly. You're right; I ignored the second term.