Question

During a softball game, a batter hits a ball upward from an initial height of 3 feet. the height, in feet, of the softball is given by s(t)=-16t^2+70t+3, where t is time in seconds and t greater than or equal to 0. which is closest to the time when the softball will be 50 feet above the ground?

Answer 1

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

$t=\frac{70\pm \sqrt{70^2-4(16)(47)}}{2(16)}$

This gives two values of time: 3.55 s and 0.83 s

Answer 2

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:

$t = \frac{-70 +/- \sqrt{70^2 - (4)*(-16)*(-47)} }{-2*16} = \frac{-70 +/- \sqrt{1892} }{-32}$

$\frac{-70 +/- \sqrt{1892} }{-32} = \frac{-70 +/- 43.5}{-32}$

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.