Answer:
2.15 seconds
Explanation:
Given the equation of the height (in feet) of the ball above the ground t seconds after a shot is released from your hand modeled by the function
f(t)= -16t² + 32t + 5.2 where t is in seconds.
Note that the value of the height at the point the ball is released from the hand is 0 feet.
To get value of t when the ball is released from your hand, we will substitute f(t) = 0 into the modeled equation as shown;
f(t)= -16t² + 32t + 5.2
-16t² + 32t + 5.2 = 0
Multiply through by 10
-160t² + 320t + 52 = 0
Dividing through by -4
40t² - 80t - 13 = 0
t = 80±√80²-4(40)(-13)/2(40)
t = 80±√6400+2080/80
t = 80±√8480/80
t = 80±92.09/80
t = 80+92.09/80
t = 172.09/80
t = 2.15 secs
Hence, the value of t when the ball is released from your hand is approximately 2.15seconds.