# Projectile Velocity

1. Jun 4, 2007

### scw287

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 66.4° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 20.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

I'm really stuck on this question not positive on my work
-4.9 t^2 +75 t sin 66.4 +0
horizontal component 75*cos(66.4) and a vertical component (75*sin(66.4))

I don't know how you would find time

2. Jun 4, 2007

### Staff: Mentor

Consider the horizontal motion. You have the distance and velocity.

3. Jun 4, 2007

### scw287

I know it relates to time....but im not sure how to tie the two together

4. Jun 4, 2007

### Staff: Mentor

What's the simplest relationship between velocity, distance, and time?

What's the difference between vertical and horizontal motion for a projectile?

5. Jun 4, 2007

### scw287

v=change x/change of time
the horizontal component doesn't have acceleration?

6. Jun 4, 2007

### Staff: Mentor

Exactly right on both. The horizontal component is constant speed motion: Distance = speed X time. What's the horizontal speed and distance traveled?

7. Jun 4, 2007

### scw287

Horizontal speed:750m/s
Distance traveled:
Distance=speedXtime
29cos66.4=750m/sXtime<<?

8. Jun 4, 2007

### Staff: Mentor

The total velocity is only 75 m/s--find the x-component of that.
is given--no calculation needed.

9. Jun 4, 2007

### scw287

oops i meant 75.0 m/s, would it be 29.0

10. Jun 4, 2007

### Staff: Mentor

That's close, but with a caculator you should be more accurate.
V_x = 75.0 m/s * cos(66.4 degrees) = ?

11. Jun 4, 2007

### scw287

30.0262774418

12. Jun 4, 2007

### scw287

How does the 11.0 m wall come into play do i have to subtract it from 30.0262

13. Jun 4, 2007

### Delta

Not yet, you've so far got the horizontal velocity (30.02). We're on to finding the time it takes to travel 20.0m (horizontally).

Then we can start on the vertical components.

14. Jun 4, 2007

### scw287

so I would just multiply 30.02m/s*20m

15. Jun 4, 2007

### scw287

V_y=75m/s*sin(66.4)
then I use the formula (V_y*t-9.80*t^2)/2
then subtract that answer from 11m right?<<<am I'm on the right track I just keep getting a very large numbers for answers

16. Jun 4, 2007

### Delta

Not quite. Remember the relationship between velocity, time and distance?

17. Jun 4, 2007

### scw287

Well i think i messed up on the time it should be .666 i think b/c of the formula v=change of x/change in time?

18. Jun 4, 2007

### scw287

so if i calculated v_y right and t right my equation would be 68.727*.666+(-9.80m/s^2*(.666)^2)/(2)

19. Jun 4, 2007

### scw287

YES I GOT IT THAT TIME!! so i guess i have learned something today