Calculating Projectile Velocity to Clear 11m Wall

[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

 

[Doc Al]

I don't know how you would find time

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

 

[scw287]

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

 

[Doc Al]

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

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

 

[scw287]

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

 

[Doc Al]

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

 

[scw287]

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

 

[Doc Al]

Horizontal speed:750m/s

The total velocity is only 75 m/s--find the x-component of that.

Distance traveled:

is given--no calculation needed.

 

[scw287]

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

 

[Doc Al]

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

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

 

[scw287]

30.0262774418

 

[scw287]

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

 

[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.

 

[scw287]

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

 

[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

 

[Delta]

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

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

 

[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?

 

[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)

 

[scw287]

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