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Math(Please help)

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Projectile motion: Let's suppose you throw a ball straight up with an initial speed of 50 feet per second from a height of feet.

a) Find the equation that describes the motion as a function of time.

My answer was x = 0 and y=-16t^2+50t+6

b) How long is the ball in the air?

This is the part I need help on. Can I put the y equation into my graphing calculator to solve for t?

c) Determine when the ball is at maximum height. Find the maximum height.

  • Math(Please help) - ,

    when is y = 0 (hits ground)?
    0 = 16 t^2 -50 t - 6 (I multiplied both sides by -1)
    0 = 8 t^2 -25 t -3
    solve quadratic for t
    t = [ 25 +/- sqrt (625 + 96) ] / 32
    t = [25 +/- 26.8514]/16
    use the + root the - root is before you threw it
    t = 3.24 seconds

    max height when velocity = 0
    V = Vo -32 t
    32 t = 50
    t = 1.56 seconds

    height = 6 + 50 (1.56) - 16 (1.56^2)

  • Math(Please help) - ,

    Thank You!!

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