A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=80t−16t^2.

Question

A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=80t−16t^2.

(a) At what time t will the ball strike the ground?
(b) For what time t is the ball more than 64 feet above the ground?

Answer 1

(a) To find the time at which the ball will strike the ground, we need to find the time when s = 0.

Setting s = 0 in the equation s = 80t - 16t^2:

0 = 80t - 16t^2
0 = t(80 - 16t)
0 = t(5 - t)(16 - t)

So, t = 0, 5, or 16.

Since the ball is thrown vertically upward, the time at t = 0 refers to the initial position when the ball is thrown. So the ball will strike the ground when t = 5 seconds.

(b) To find the time when the ball is more than 64 feet above the ground, we need to find when s > 64.

Setting s > 64 in the equation s = 80t - 16t^2:

80t - 16t^2 > 64
16t^2 - 80t + 64 < 0
16(t^2 - 5t + 4) < 0
16(t - 4)(t - 1) < 0

So, 1 < t < 4 is the time when the ball is more than 64 feet above the ground.

A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s​ (in feet) of the ball from the ground after t seconds is s=80t−16t^2.

(a) To find the time at which the ball will strike the ground, we need to find the time when s = 0.

A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=80t−16t^2.