Question

"A ball is thrown vertically upward from the top of a building 64 feet tall with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is given by s(t)=64+48t−16t². After how many seconds does the ball strike the ground? After how many seconds will the ball pass the top of the building on its way down?

a) After 2 seconds, the ball passes the top of the building.
b) After 3 seconds, the ball passes the top of the building.
c) After 5 seconds, the ball passes the top of the building."
d) none of these.

Answer 1

The ball strikes the ground after 3 seconds and passes the top of the building on its way down immediately after it is thrown.

So option (B) is the correct answer

Step-by-step explanation:

The equation representing the distance of the ball from the ground is given by s(t) = 64 + 48t - 16t². To find the time at which the ball strikes the ground, we need to find the value of t when s(t) = 0. So, we need to solve the equation 64 + 48t - 16t² = 0.

Using the quadratic formula, we have t = (-48 ± sqrt(48² - 4(-16)(64)))/(2(-16)).

Calculating this, we get t = (-48 ± 16 sqrt(10))/(-32). Evaluating the positive root, we find t ≈ 3 seconds. Therefore, the ball strikes the ground after 3 seconds. To find the time at which the ball passes the top of the building on its way down, we can set s(t) equal to the height of the building (64 feet) and solve for t. So, we need to solve the equation 64 + 48t - 16t² = 64.

Again using the quadratic formula, we find t = (-48 ± sqrt(48² - 4(-16)(0)))/(2(-16)). Simplifying, we get t = (-48 ± sqrt(48²))/(2(-16)). Calculating this, we get t = (-48 ± 48)/(2(-16)). Evaluating the positive root, we find t = 0 seconds. Therefore, the ball passes the top of the building on its way down immediately after it is thrown.