Question

A ball is thrown into the air with an initial velocity of 32 feet per second, from a height of 11 feet. The equation for its height time t is given by v(t)=-16t2 + 32t+11. After how many seconds will the ball hit the ground (round your answer to the neare tenth of a second)?

Answer 1

okay so..

v(t) = -16t^2 + 32t + 11
0 = -16t^2 + 32t + 11

We can solve for t using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

Where a = -16, b = 32, and c = 11.

Plugging in these values, we get:

t = (-32 ± sqrt(32^2 - 4(-16)(11))) / 2(-16)
t = (-32 ± sqrt(1024 + 704)) / (-32)
t = (-32 ± sqrt(1728)) / (-32)

We can simplify the square root of 1728 to 24 times the square root of 3:

t = (-32 ± 24sqrt(3)) / (-32)

Simplifying further, we get:

t = 1 ± 3/2 sqrt(3)

We take the positive root, since we're looking for a time, so we get:

t = 1 + 3/2 sqrt(3) ≈ 2.4 seconds

So the ball will hit the ground after about 2.4 seconds.