Question

mary pitches a ball. the initial height is 4 ft and velocity 55 ft per second. when will the ball hit the ground?

Answer 1

Answer:

t = 3.5 sec

Step-by-step explanation:

The equation that represents the position of the ball as a function of time is the following:

$h(t) =P_{o} + V_{o}t+\frac{1}{2}2at^2$

Where:

$P_{o}$ = Initial position of the ball (4 feet tall)

$V_{0}$ = Initial speed of the ball (55feet / s)

a = acceleration (9.8 $\frac{m}{s^2}$, or in this case 32.16$\frac{ft}{s^2}$)

t = time in seconds

Then we want to know how long it takes to get to the ground, for this we equal h (t) = 0 and clear t.

So:

$0 = 4 + 55t -16.08t ^ 2$

Solving the second degree equation we have:

$\frac{-b+\sqrt{b ^ 2-4ac}}{2a} \\ \frac{-b-\sqrt{b ^ 2-4ac}}{2a}$

$\frac{-55+\sqrt{55^2-4 (-16.08) 4}}{2 (-16.08)}= -0.0712\\\frac{-55-\sqrt{55^2-4 (-16.08) 4}}{2 (-16.08)} = 3.492$

t = -0.071 sec

t = 3.5 sec

We must take the positive solution.

So:

t = 3.5 sec