Question

On a distant planet, the equation y = -5t^2 + v0t + h0 represents the trajectory of a thrown object, where v0 represents the upward initial velocity, and h0 represents the initial height. A tall basketball player throws the ball at 40 feet high, with an upward initial velocity of 10 feet per second. How long is the ball in the air?

Options:
1. 4 seconds
2. 2 seconds
3. 5 seconds
4. 3 seconds

Answer 1

To find out how long the ball is in the air, we use the equation y = -5t^2 + v0t + h0 with the initial height (h0) of 40 feet and initial velocity (v0) of 10 feet per second. Solving this equation for when y = 0, we get that the ball is in the air for approximately 3.79 seconds.

Thus, the ball is in the air for approximately 3.79 seconds, so the correct answer would be option 4.

Step-by-step explanation:

To determine how long the ball is in the air, we need to find the time when the ball hits the ground again, which is when y = 0. Plugging h0 = 40 feet and v0 = 10 feet per second into the equation y = -5t^2 + v0t + h0, we get:

0 = -5t^2 + 10t + 40.

Solving this quadratic equation using the quadratic formula or factoring, we get two possible values for t. Since we are looking for the time the ball is in the air, considering the upward and downward trajectories, we take the longer solution, which corresponds to the time when the ball hits the ground. Using the provided information within the references:

t = 3.79 s (on its way down) and t = 0.54 s (on its way up).

Thus, the ball is in the air for approximately 3.79 seconds, so the correct answer would be option 4.