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) = -16t^2+vt+s

Answer 1

Answer:

4.643 seconds

Step-by-step explanation:

We have been given that

s = 6 feet

v = 73 feet per second

Substituting these values in the formula $H(t)=-16t^2+vt+s$

$H(t)=-16t^2+73t+6$

When the ball hits the ground, the height becomes zero. Thus, H(t)=0

$-16t^2+73t+6=0$

We solve the equation using quadratic formula $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values $a=-16,\:b=73,\:c=6$

$t_{1,\:2}=\frac{-73\pm \sqrt{73^2-4\left(-16\right)6}}{2\left(-16\right)}\\\\=\frac{-73\pm\sqrt{5713}}{-2\cdot \:16}\\\\=-0.081,4.643$

time can't be negative. Hence, t = 4.643.

Hence, the ball will take  4.643 seconds to hit the ground.

Answer 2

Hey! The correct time is 4.64 seconds.