Question

A ball is thrown vertically in the air with a velocity of 95ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 120ft. Round your answer(s) to the nearest tenth of a second.

Answer 1

Answer:

$t_1 = 1.8\ s\\\\t_2 = 4.1\ s$

Step-by-step explanation:

If the equation $h = -16t^2 + v_0t$ represents the position of the ball as a function of time then, to find in which second the ball reaches 120 feet must substitute $h = 120$ in the equation of the height and solve for t.

$120 = -16t ^ 2 + v_0t$

If the initial velocity is 95 feet/s then $v_0 = 95$

Then:

$120 = -16t ^ 2 + 95t\\\\-16t ^ 2 + 95t -120 = 0$

Use the quadratic formula

$t_1 = \frac{-b+\sqrt{b^2 -4ac}}{2a}\\\\t_2 = \frac{-b-\sqrt{b^2 -4ac}}{2a}$

Where, for this problem:

$a = -16\\b = 95\\c = -120$

So

$t_1 =\frac{-95+\sqrt{(95)^2 -4(-16)(-120)}}{2(-16)} = 1.8\ s\\\\t_2=\frac{-95+\sqrt{(95)^2 -4(-16)(-120)}}{2(-16)} = 4.1\ s$

This result means that the ball reaches 120 feet for the first time at 1.8 seconds, then begins to descend and on its descent again reaches 120 feet at t = 4.1 seconds.

Answer 2

Answer:

The Answer Was "To the nearest tenth of a second, the ball was in the air for 2.4s"

Step-by-step explanation: