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Physics I

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A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 61.2° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 20.5 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?

  • Physics I -

    We will neglect air resistance and propulsion of the rocket, and treat the rocket as a projectile.

    v0=initial velocity = 75 m s-1
    θ=angle with horizontal

    horizontal component of velocity, vh
    =v0cos θ
    Distance from the wall, D = 20.5 m
    Time to reach the wall, t1
    = D/vh

    Vertical component of initial velocity
    = vv
    = v sin θ

    Height h of rocket after time t1
    = vv*t1 -g (t1)²

    Clearance from the wall, C
    = h - 11m

    I get about 23m.

  • Physics I -

    23m is wrong

  • Physics I-correction -

    Thank you Brittany, there was a factor (1/2) missing near the end. Here is the corrected calculation.

    Height h of rocket after time t1
    = vv*t1 -(1/2)g(t1)²

    Clearance from the wall, C
    = h - 11m

    I get a little less than 25m using g=9.8 m s-2.

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