Question

A ball is thrown vertically upward from the top of a building with an initial velocity of 96 feet per second. The ball's h height after t seconds is given by the equation h = −16t2 + 96t + 100. How long will it take for the ball to hit the ground? (to the nearest tenth of a second)
A) 5.9 seconds
B) 6.3 seconds
C) 6.9 seconds
D) 7.2 seconds

Answer 1

Answer:

C

Step-by-step explanation:

The ball's h height after t seconds is given by the equation

$h =-16t^2 + 96t + 100.$

When the ball hits the ground, the height is equal to 0, so substitute h = 0 into the function expression and solve it for t:

$-16t^2+96t+100=0\ \ [\text{Divide by -4}]\\ \\4t^2-24t-25=0\ \ [\text{In this quadratic equation }a=4,\ b=-24,\ c=-25]\\ \\D=b^2-4ac=(-24)^2-4\cdot 4\cdot (-25)=576+400=976\\ \\t_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-(-24)\pm\sqrt{976}}{2\cdot 4}=\dfrac{24\pm4\sqrt{61}}{8}\approx -0.9,\ 6.9$

Since time cannot be negative, the ball will hit the ground in 6.9 seconds

Answer 2

Answer:

C

Step-by-step explanation:

The ball's h height after t seconds is given by the equation

When the ball hits the ground, the height is equal to 0, so substitute h = 0 into the function expression and solve it for t: