A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is S(t)=96t-16t^2. At what time t will the ball strike the ground? For what time t is the ball more than 96 feet above the ground?
final position: s(t)=0
0=96t-16t^2=16t(6-t)
so t=6 it hits the ground.
for your last part, you want
-16t^2 + 96t > 96
let's see when -16t^2 + 96t = 96
t^2 - 6t +6 = 0
t = (6 ± √12)/2
= 3 ± √3
= 1.268 or 4.732
so between these two times the ball is above 96 ft
so the ball is above 96 ft for 3.464 seconds
To find the time at which the ball will strike the ground, we need to find when the distance of the ball from the ground is zero. In other words, we need to solve the equation S(t) = 0.
Given that S(t) = 96t - 16t^2, we can set it equal to zero: 96t - 16t^2 = 0.
Factoring out common terms, we get: 16t(6 - t) = 0.
Now we have two possibilities:
1. 16t = 0, which means t = 0.
This solution implies that the ball is at the ground at t = 0 (initial time).
2. 6 - t = 0, which means t = 6.
This solution implies that the ball will hit the ground at t = 6 seconds.
Therefore, the ball will strike the ground at t = 6 seconds.
To find the time when the ball is more than 96 feet above the ground, we need to find when the distance of the ball from the ground is greater than 96. In other words, we need to solve the inequality S(t) > 96.
Given that S(t) = 96t - 16t^2, we can set it greater than 96: 96t - 16t^2 > 96.
Rearranging the inequality: 16t^2 - 96t + 96 < 0.
Dividing through by 16: t^2 - 6t + 6 < 0.
To solve this inequality, we can factorize it or use the quadratic formula.
Using the quadratic formula, t = (6 ± √(6^2 - 4(1)(6))) / (2(1)).
Calculating the discriminant (√(6^2 - 4(1)(6))) gives us √(36 - 24) = √12 = 2√3.
So, the solutions for t are: t = (6 + 2√3) / 2 ≈ 6 + 1.155 = 7.155 and t = (6 - 2√3) / 2 ≈ 6 - 1.155 = 4.845.
Therefore, the ball is more than 96 feet above the ground for t values between approximately 4.845 and 7.155 seconds.