How High Does the Rocket Clear the Wall?

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A rocket is launched at 75.0 m/s at a 60.0º angle, aiming to clear an 11.0m wall located 27.0m away. The initial calculations for time and vertical displacement are incorrect; the time must be calculated based on the horizontal distance to the wall. The correct approach involves determining the horizontal velocity and using it to find the time to reach the wall. Once the correct time is established, the vertical displacement can be calculated to determine how much the rocket clears the wall. The discussion emphasizes the importance of using the right formulas and understanding the relationship between horizontal and vertical motion.
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Homework Statement


A rocket is fired at a speed of 75.0 m/s from ground level at an angle of 60.0º above the horizontal. The rocket is fired toward an 11.0m high wall, which is located 27.0m away. By how much does the rocket clear the wall?


Homework Equations


y=vot-1/2gt^2

possibly x=vyt

The Attempt at a Solution



I tried using the formula listed above (first one) to find the time.

I made it so it was t= \sqrt{}2(-11)/-9.8

To get t=1.5 seconds

I then did sin60º(75) to get 64.95m

I then plugged in the numbers into the equation listed above (2nd one)

I plugged in the following:

64.95m(1.5s)

To get 97.4m, then I subtracted 11m from 97.4m, to get 86.4m

The rocket clears the wall by 86.4 meters



Is this wrong? If it is, can someone guide me in the right direction, by which formulas I should use?

Thanks so much guys, I appreciate it.






Peace
 
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Please guys, Need help as soon as possible :/
 
calculate the time taken to reach the wall... use horizontal velocity and horizontal distance = 27.0m.

using that time and vertical velocity... calculate the vertical displacement within the time you just got... then subtract the height of the wall.
 
Don't you find the maximum vertical displacement by using the formula y=vytI found the time by using the formula t=1/2gt^2

g= -9.8 m/s^2

to get time= 1.5 seconds

I then found the vertical velocity by using sine(trig)

sin60º(75) to get 64.95 m/s

Then Plugged it in

y= 64.95(1.5)

=97.4m - 11.0m = 86.4m

I said this in the first post. Is this correct?
 
No, it's not correct. Learningphysics is right, you need to calculate the time based on the horizontal length of 27m, then plug the time into the formula to get the vertical distance.
 
The time you're calculating isn't right. You're calculating the time to get to a height of 11.0m... that's not the time you need... Follow the method chocokat and I are giving. Get the time to reach the wall... what is the horizontal velocity? Horizontal displacement is 27m.

You need to find out by how much the wall is cleared... in other words you need the vertical displacement when the horizontal displacement is 27m. What is the time to get to 27m...
 
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