Algebra 2 for Precalc

Question

A baseball is hit upward at 120 feet per second. Use the formula h (t) = -16t^2+ vot, where h(t) is the height of an object in feet, vo is the object's initial velocity in feet per second, and t is the time in seconds. Ignoring the height of the ball when it was hit, how long does it take for the ball to hit the ground?

Answer 1

When I teach this topic, I like to have my students visualize an object being launched directly upwards with an initial velocity and gradually slows down due to gravity, before falling back to the ground. The -16 comes from 1/2at^2, where a is the acceleration of an object due to gravity on Earth (-32 ft/s, it's -9.8 m/s when using the metric measurements). The time to peak (if initial height is 0) is equal to the time from peak back to the ground. Example, if it takes 8 seconds to go from ground to peak, it takes 8 seconds to go from peak back to the ground. Raymond's response is probably the easiest approach to this question. However, you can also use the quadratic equation, which may be more useful for questions that aren't easy to factor.

Quadratic Equation:

(-b ± √(b^2 - 4ac))/2a

In your question, a = -16, b = 120, c = 0. Plug those values into the equation.

-120 ± √(120^2 - 4(-16)(0)))/2(-16) = (-120 ± 120)/-32, split the ± into two separate equations.

(-120 + 120) / -32 = 0 / 32 = 0 seconds ::::: 1 solution

(-120 - 120) / -32 = -240 / -32 = 7.5 ::::: another solution

Answer 2

-16t^2 +120t =0

factor

-8t(2t - 15) = 0

t=0 or 15/2 = 7.5 seconds to hit ground