# Homework Help: Review Q #6 (Projectile Motion)

1. Sep 21, 2010

### TeenieWeenie

Thanks Thaakisfox! Problem solved!

1. The problem statement, all variables and given/known data
A track athlete leaves the ground with initial speed 10.2 m/s and makes an 8.8 meter long jump. Calculate the most likely angle of the direction of his take off with respect to the horizontal.

Vo = 10.2 m/s
x = 8.8 m
xo = 0 m
Ao = ?
2. Relevant equations
x=Vo * CosAo * t
Vx = VoCosAo

3. The attempt at a solution
I have a missing t (time) variable or I'm using the wrong formula(s)?

Last edited: Sep 21, 2010
2. Sep 21, 2010

### Thaakisfox

Here you have to calculate a maximum. . Most probably the athlete jumped so that the distance was maximal. I would take the maximal height of the jump to be given and maximize with respect to the distance.

3. Sep 21, 2010

### TeenieWeenie

R= [Vo^2 * sin2Ao] / g
where R = range = x = 8.8m

8 = [(10.2^2)*sin2Ao]/ 9.8m/s

8/9.8 = 10.2^2 *sin2Ao

[8/9.8]/(10.2^2) = sin2Ao

(sin2Ao = 2sinAocosAo ?)

[(8/9.8)/(10.2^2)]/2 = sinAoCosAo ?

Um.....I don't know what to do from there...?

or if h was the given maximum height...
8 = [(10.2^2)sinAo] / 2(9.8)
[8/2(9.8)] = (10.2^2)sinAo
[8/19.6]/(10.2^2) = sinAo
sin-1[8/19.6]/(10.2^2) = Ao
.0039231 = Ao.

Errr.....error?

4. Sep 21, 2010

### Thaakisfox

Yes. Thats the range of the projectile. So this has to be maximal. Now R is given, Vo is given, g is given, hence the only thing that can change is Ao the angle of the jump.
So we need sin(2Ao)= maximal. we know that the max of sin is 1. Hence sin(2Ao)=1.

Now at what angle does this occur?

5. Sep 21, 2010

### TeenieWeenie

R= [Vo^2 * sin2Ao] / g
was the key to solving this problem then?
R = range of projectile = maximal = 8.
sin(2Ao) = maximal and will never be greater than 1 so it has to be = 1 which is 90 degrees or 2 * 45 degrees ?

Could you explain that a little more?

6. Sep 21, 2010

### Thaakisfox

Yes thats it. Exactly.
The athlete tries to jump, so that his/her (his in this case as female athletes usually dont jump above 7.5 range is maximal. Now we have an expression for the range as a function of the velocity and the angle. We know that the jump was 8m, we also somehow know that the velocity was 10.2. Since we know that the range was maximal, and we know the velocity, we only have to figure out at what angle this could have happened.
And that is the sine should be 1. Which means that the angle of the jump is 45 degrees.
Its actually a good thing to remember that when you throw something (stone or anything messing around) the best is to throw it at a 45 degree angle with the horizontal because then it will go the furthest.

7. Sep 21, 2010

### TeenieWeenie

OOO! I understand now!
That should have been a no-brainer :|

Thanks Thaakisfox! Problem solved!