Mathematics: Aball 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.

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Aball is thrown vertically in the air with a velocity

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kayleerichards666

06.01.2022

Mathematics

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

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:

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

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

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.

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Aball 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.

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