Question

Amir pitches a baseball at an initial height of 6 feet, with a velocity of 73 feet per second. If the batter misses, about how long does it take the ball hit the ground?
Hint: Use H(t) = −16t2 + vt + s.

Answer (1) 4.64 seconds

Answer (2) 2.94 seconds

Answer (3) 2.28 seconds

Answer (4) 0.08 second

Answer 1

Answer:

Option (1). 4.64 seconds.

Step-by-step explanation:

It is given in the question that a baseball pitcher was at an initial height of 6 feet.

Velocity of the ball is 73 feet per second.

The expression that represents the event is

H(t) = -16t² + vt + s

Here s = initial height = 6 ft.

H(t) = 0

velocity of the ball = 73 feet per second

The expression will be

-16t² + 73t + 6 = 0

Now we will get the solution for t.

$t=\frac{-73\pm \sqrt{(73)^{2}+4\times 16\times 6}}{2(-16)}$

$=\frac{-73\pm \sqrt{5329+384}}{-(32)}=\frac{-73\pm 75.58}{-32}$

$t=-.081$ and $t=4.64$

Time can not be negative so t = 4.64 sec is the answer.

Option 1). 4.64 is the answer.

Answer 2

The solution would be like this for this specific problem:

x = [ -73 +/- sqrt (73^2 – 4 * -16 * 6) ] / (2 * -16)

73^2 - 4*-16*6 = 5713

sqrt 5713 = 75.58

(-73 - 75.58) / -32 = 4.64