Question

The height in feet of an object projected upward as a function of the elapsed time in seconds since it was launched can be represented by the formula shown below: h(t) = -16t^2 + v0t + h0. In the formula, h(t) is height in feet; t is time in seconds; v0 is the initial velocity; and h0 is the initial height. An object is projected upward with an initial velocity of 50 feet per second from a starting height of 9 feet. When will the object hit the ground?

a) t = 3 seconds
b) t = 2.5 seconds
c) t = 4 seconds
d) t = 5 seconds

Answer 1

The object will hit the ground at t = 3.79 seconds.

Step-by-step explanation:

The height of an object projected upward can be represented by the formula h(t) = -16t^2 + v0t + h0, where h(t) is the height in feet, t is time in seconds, v0 is the initial velocity, and h0 is the initial height. In this case, the object is projected upward with an initial velocity of 50 feet per second from a starting height of 9 feet. We want to find out when the object will hit the ground.

To find the time when the object hits the ground, we need to set h(t) equal to zero and solve for t. Using the quadratic formula, we get two solutions: t = 3.79 seconds and t = 0.54 seconds. Since the ball is at a height of 10 feet at two times during its trajectory (once on the way up and once on the way down), we take the longer solution, which is t = 3.79 seconds. Therefore, the object will hit the ground at t = 3.79 seconds.