1. Jan 31, 2008

jj8890

[SOLVED] Projectile motion question

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

A 2.291m tall man shoots a ball at 19m/s at an angle of 74 degrees above the horizontal into a hoop that is 3.048m tall, how long does it take the ball to reach its max height? To reach the hoop? What is the horizontal length of the shot?

2. Relevant equations
I used Y=Voy-.5gt^2 to find the max height but not sure if this is the rights equation

3. The attempt at a solution

I used Y=19t -.5(9.81)t^2 to get t~1.937 s

I am confused because the start and end heights are different (the man is 2.291, tall and the hoops is 3.048m). I'm not sure how to deal with this.

2. Jan 31, 2008

rocomath

Take y initial to be the man's height and the hoop to be y final. (if i'm not mistaken)

Also, what is the velocity at max height?

You'll need these 5 equations (one which is already simplified, plugged in the 1st into the 2nd eliminating time)...

$$x=(v_0\cos\theta)t$$

$$\Delta y=(v_0\sin\theta)t-\frac 1 2 gt^2$$

$$\Delta y = x\tan\theta - \frac {gx^2}{2(v_0\cos\theta)^2}$$

$$v_x=v_0\cos\theta$$

$$v_y=v_0\sin\theta-gt$$

Last edited: Jan 31, 2008
3. Jan 31, 2008

jj8890

Thanks for the help though I greatly appreciate it. The velocity at the max height would be zero right? I used the first equation and got ~1.89s so this would be the time for the max height? Would I just double that to get the time when it goes through the hoop? I am still unsure where to use the other two equations.

4. Jan 31, 2008

rocomath

Well, since we don't have the time it takes for the ball to reach the hoop. We need to find the horizontal distance that it travels.

Use the 3rd equation that I provided and solve for x. Use the quadratic equation. Then plug that x back into the 1st equation to solve for t.

Also, do you have the answer to this question? I don't really like physics very much, but I'll try my best to help.

5. Jan 31, 2008

jj8890

Ok...I used the 3rd equation and solve for x and got x~15.95m. This is the horizontal distance? I then plugged that x back into the 1st equation and got t~3.045s. Which would be the time that it took to go the distance? I'm just trying to make sure i'm doing this right. No, I do not have the answer...Thanks for your help again.

6. Jan 31, 2008

rocomath

My equation reduces to ...

$$.1788362932x^2-3.487414444x+.757=0$$

x = 19.44617835 m

t = 3.713150811 s

Last edited: Jan 31, 2008
7. Jan 31, 2008

jj8890

Ok...I see what i did wrong...must be getting tired...thanks..so the 20.53 is the distance and I got ~3.92s fo the time. This would be the time for the whole trip right? About the max height, what equation would I use for that or was it correct to begin with?

8. Jan 31, 2008

rocomath

Check my equation again on Post #6, I updated it b/c I kept calculating initial velocity as 19.6m/s, hehe I'm tired too sorry.

And to find the max height, velocity would be 0 as you stated. So use the 5th equation to find the time.

t = 1.861770869 s

9. Jan 31, 2008

jj8890

Thank you so much. You were a great help.

10. Jan 31, 2008

rocomath

Make sure to check this later. If I was wrong on my approach, someone will definitely have yelled and corrected me :-]