Answer: 3.55 s and 0.83 s
Explanation:
The height (in feet) of the softball is given by:
s(t)=-16t²+70t+3
we need to find the time when the ball is 50 feet above the ground:
⇒50 = -16t²+70t+3
we need to solve the quadratic equation
⇒16t²-70t+47 = 0
[tex]t=\frac{70\pm \sqrt{70^2-4(16)(47)}}{2(16)}[/tex]
This gives two values of time: 3.55 s and 0.83 s
Answer: The solutions are:
t1 = (-70 + 43.5)/-32 = 0.83 seconds
t2 = (-70 - 43.5)/-32 = 3.55 seconds
Explanation: Here the equation for the height of the ball is s(t) = -16t^2+70t+3
If we want to find the time where the ball is at 50 ft above the ground, we need to solve the equation:
s(t) = 50 = -16t^2+70t+3
0 = -16t^2+70t+3 - 50 = -16t^2+70t - 47
now, the solutions are:
[tex]t = \frac{-70 +/- \sqrt{70^2 - (4)*(-16)*(-47)} }{-2*16} = \frac{-70 +/- \sqrt{1892} }{-32}[/tex]
[tex]\frac{-70 +/- \sqrt{1892} }{-32} = \frac{-70 +/- 43.5}{-32}[/tex]
So the solutions are:
t1 = (-70 + 43.5)/-32 = 0.83 seconds
t2 = (-70 - 43.5)/-32 = 3.55 seconds
Where we can assume that t1 is when the ball is going up, and t2 is when the ball already reached is max point and started to fall to the ground.
