Question

The height of an object after it is released can be modeled by the function f(t) = -16t^2 + vt + s, where t is the number of seconds after the object is released, v is the upward speed at release, and s is the starting heightIf a quarterback throws a ball from his hand 6 feet in the air at a speed of 25 feet per second, how much time does his teammate have to catch the ball?

Answer 1

The teammate of the quarterback has 1.774 seconds to catch the ball.

Explanation:

Given the height of the object to be:

f(t) = -16t² + vt + s

Given that the quarterback throws the ball in his hand 6 fts in the air, at a speed of 25ft/s.

We want to find how much time his teammate have to catch the ball.

This is the same as finding the value of f(t) = 0

That is

-16t² + vt + s = 0

With s = 6ft, v = 25 ft/s

So, we solve

-16t² + 25t + 6 = 0

Using the quadratic formula

t = [-b ± √(b² - 4ac)]/2a

Where a = -16, b = 25, and c = 6

t = [-25 ± √(25² - 4×(-16)×6)]/2(-16)

= 25/32 ± √(625 + 384)/(-32)

= 25/32 ± (-1/32)√1009

= 25/32 ± (-0.993)

= 0.781 ± 0.993

t = 0.781 + 0.993

= 1.774

Or

t = 0.781 - 0.993

= -0.212

We take the positive value for time, which is t = 1.774s

Answer 2

1.7739 seconds

Explanation:

In the problem, v = 25 and s = 6, (we are counting height in feets). We want t such that f(t) = 0. We can solve it with the quadratic formula

`r_1,r_2 = (-b ^+_- \, √(b^2-4ac) )/(2a)`

Here, a = -16, b = 25 and s = 6, thsu

`r_1,r_2 = (-25 ^+_- \, √(25^2-4(-16)6) )/(2(-16)) = (-25 ^+_- 31.76)/(-32)`

We take only the - before the square root because if we sum we would obtain a negative root. Thus, r = (-25-31.76)/-32 = 1.7739.

His teammate have to catch the ball in the second 1.7739