Projectile velocity....still don't understand this problem?

Question

scw287

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Projectile Velocity
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 know you use V_x = 75.0 m/s * cos(66.4 degrees) =30.026

not sure what you do next

Answer 1

You're off to a good start. Here's what I would do next:

1. Calclulate how long it takes for the rocket to reach the wall (hint: you've already got the numbers. The wall is 20.0 m away, and the rocket's horizontal speed is 30.026 m/sec. How long does it take to go 20 m at that speed?) Calculate that time, and call it "t".

2. Calculate how high the rocket will be after "t" seconds. You should know that the formula for this is:

h = V_y * t - (g*t^2)/2

Since you got V_x correct, I'm assuming you also know how to calculate V_y. Then you just plug in V_y, plus the "t" you calculated in Step 1, into the above formula to get h.

3. The h you just calculated is the rocket's height above the _ground_ at the time it reaches the wall. But you want to know its height above the top of the _wall_. That's a simple subtraction problem (remember the wall is 11.0 m high).

Hope this helps!

Answer 2

be conscious that each and each projectile formula includes 3 variables (actually 4, yet acceleration is considered consistent in projectile issues). you purely could use the formula the place you could in high-quality condition 2 of the given values and discover the relax variable. in this project, you have the preliminary speed with the perspective, extremely some the distances, and the time on the main suitable of the wall. a) vertical distance while the ball is rapidly above the wall: you would be utilising t=2.2s through fact its the time of the stated requirement; and V0= 18.18m/s ? = fifty 3.0° first get the y-part of the fee through fact which you're finding the y-distance: Vy = 18.18 sin fifty 3 = 14.fifty two m/s the equation that includes (t,y,Vy) is: y = Vy*t + ½*a*t² = [14.fifty two*2.2] + [½*(-9.eighty one)*2.2²] = 8.2038m b) horizontal distance: of the easy formula, you purely have a million equation that includes the x distance, it incredibly is: x = Vx*t; there are derived formula, yet i don't propose utilising it, that's proper to appreciate the basics first until now utilising complicated equations. needless to say, u purely have Vx, u nonetheless could seem for t, you could not use 2.2s through fact its the time at real of the wall, not on the playground itself. The available values are the y-distance of the playground, and preliminary y-speed. subsequently, you would be utilising a formula that includes (y,Vy,t), then remedy for t: y = Vy*t + ½*a*t² 4.8 = 14.fifty two*t + ½*(-9.eighty one)*t² t² - 2.ninety six*t + 0.ninety 8 = 0 <-- quadratic, it incredibly is through fact the ball passes the 4.8 distance two times; until now it clears the wall and the different is at its touchdown. you could decide for the main present day time. t = 0.38s; 2.58s then we use 2.58s and Vx and remedy for x: Vx = 18.18 cos fifty 3 = 10.ninety 4 X = 10.ninety 4 * 2.fifty 8 = 28.2252m <--it incredibly is the gap from the factor of launch; subtract the gap from the launch factor to the wall to get the gap of the ball from the wall 28.2252 - 24 = 4.2252m from the wall **i'm hoping u understand :)

Answer 3

use vx to find the time it takes to reach the wall.

use use the equation

y=0.5at^2 +vt

use vy for the intial v (use sin to find vy). 'a' should be 9.8. Plug in the time you found in the first part and this will give you the height of the rocket when it reaches the wall. Then subtract the height of the wall, to find the clearance.