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7 January, 08:49

The formula for height, in feet of a projectile under the influence of gravity is given by h=-16t^2+vt+s, where t is the time in seconds, v is the upward velocity at the start, and s is the starting height. Marvin throws a baseball straight up into the air at 70 feet per second. The ball leaves his hand at a height of 5 feet. When does the ball reach a height of 75 feet?

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  1. 7 January, 09:11
    0
    1.55s

    Step-by-step explanation:

    The height equation of the projectile is given as;

    h = - 16t^2 + vt + s ... 1

    where;

    t is the time in seconds,

    v is the upward velocity at the start,

    s is the starting height.

    Given;

    v = 70 ft/s

    s = 5 ft

    Substituting the values equation 1 becomes;

    h = - 16t^2 + 70t + 5

    For the ball to reach 75 ft, h = 0

    75 = - 16t^2 + 70t + 5

    -16t^2 + 70t + 5 - 75 = 0

    -16t^2 + 70t - 70 = 0

    solving the quadratic equation;

    Time t = 2.83 s and t = 1.55s

    Therefore, the ball will first reach 75 ft after 1.55 seconds.
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