Question

WHOEVER IS CORRECT GETS 55 POINTS!!! PLZ HELP FASTTTTT!!!!!!!martin throws a ball straight up in the air. the equation h(t)=-16t^2+40t+5 gives the height of the ball, in feet, t seconds after martin releses it. how many seconds before the ball martin threw hits the ground? AND SHOWWW WORKK

Answer 1

Given that Martin throws a ball straight up in the air.

The equation $h(t)=-16t^2+40t+5$ gives the height of the ball, in feet, t seconds after martin releases it.

We need to determine the time that it takes the ball to hit the ground.

Time taken:

To determine the time 't', let us equate h(t) = 0 in the equation $h(t)=-16t^2+40t+5$

Thus, we have;

$0=-16t^2+40t+5$

Switch sides, we get;

$-16 t^{2}+40 t+5=0$

Now, we shall solve the equation using the quadratic formula.

Thus, we have;

$t=\frac{-40 \pm \sqrt{40^{2}-4(-16) 5}}{2(-16)}$

Solving, we get,

$t=\frac{-40 \pm \sqrt{1600+320}}{-32}$

$t=\frac{-40 \pm \sqrt{1920}}{-32}$

$t=\frac{-40 \pm 8\sqrt{30}}{-32}$

$t=\frac{8(-5 \pm \sqrt{30})}{-32}$

Cancelling the common terms, we get,

$t=\frac{-5 \pm \sqrt{30}}{-4}$

Thus, the roots of the equation $t=\frac{-5 + \sqrt{30}}{-4}$ and $t=\frac{-5 - \sqrt{30}}{-4}$

Simplifying the roots, we get,

$t=\frac{-5 +5.477}{-4}$   and   $t=\frac{-5 -5.477}{-4}$

$t=-0.119$   and   $t=2.619$

Since, t cannot take negative values, then $t=2.619$

Hence, It takes 2.6 seconds for the ball to hit the ground.