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Josh fires a catapult 18 ft above the ground with an initial velocity of 25 ft/sec. The misslie leaves the catapult at an angle of degree theta with the hoizontal and heads toward a 40 ft wall 500 ft from the launch site. If the missle is fired at a 20 degree angle, does it clear the wall? Justify the answer

If the missle is fired at a 35 degree angle, how high does it go?

  • pre-calculus -

    The vertical component of the velocity is 25sin20° = 8.55

    So, the height y is
    y = 18 + 8.55t - 16t^2
    If the missile just clears the wall, then
    40 = 18 + 8.55t - 16t^2
    Hmm. No solutions.

    Check for typos, and take a look at wikipedia's article on "trajectory" for details of the range, height, etc.

  • pre-calculus -

    initial horizontal velocity: 25*cos(theta)=v1
    initial vertical velocity:

    where theta is the angle to the horizontal.

    max height:

    will it go over the wall:
    v1*t=500 ft

    solve the above equation for t and plug into:
    18+v2*t-(32ft/s^2)/2*t^2= height at given time

  • pre-calculus -

    one correction:

    "max height:
    (v2)^2-2*32ft/s^2*(height)=0" is wrong

    it should be:
    (v2)^2-2*32ft/s^2*(height -18)=0

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