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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

User Marczellm
by
9.2k points

2 Answers

4 votes

Answer:

C

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:

User Ahbou
by
8.0k points
1 vote

Answer:

C

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)=(-b\pm√(D))/(2a)=(-(-24)\pm√(976))/(2\cdot 4)=(24\pm4√(61))/(8)\approx -0.9,\ 6.9

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

User Xersiee
by
8.3k points
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