Question

while serving the ball at a beach volleyball game, a player tosses the ball up from a height of 4 feet at a velocity of 18 feet per second. after it starts to fall back down, the player jumps up and hits the ball at a height of 7 feet. to the nearest tenth of a second, how long is the ball in the air before the player hits it? hint: use the formula h = -16t² + vt + s.

Explanation:

Step1: Substitute values into formula

$7 = -16t^2 + 18t + 4$

Step2: Rearrange to quadratic form

$16t^2 - 18t + 3 = 0$

Step3: Apply quadratic formula $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$a=16, b=-18, c=3$; $t=\frac{18\pm\sqrt{324-192}}{32}=\frac{18\pm\sqrt{132}}{32}$

Step4: Calculate solutions, take larger t (falling)

$t\approx\frac{18+11.489}{32}\approx0.921$

Step5: Round to nearest tenth

$t\approx0.9$

Answer:

0.9 seconds