Question

A volleyball player serves the ball from a height of 6.5 feet above the ground with an initial vertical velocity of 21 feet per second. Which function models the height h (in feet) of the ball t seconds after it is served?
A) h = 21t ­- t^2 + 6.5
B) h = 16t^2 + 6.5
C) h = 16t^2 + 21t + 6.5
D) h = 6.5 + 21t ­ 16t^2

Answer 1

Answer:

Option D. $H(t) = 6.5+21t-16t^2$

Step-by-step explanation:

we know that

If air resistance is ignored, the height h (in feet) of the ball t seconds after it is served is given by

$H(t) = - \frac{1}{2}g t^2 + V_0t + H_0$

where

g is the acceleration due to gravity which on earth is approximately equal to 32 feet / sec^2

V_0 is the initial velocity (when t = 0 )

H_0 is the initial height (when t = 0)

In this problem we have

$V_0=21\ ft/sec\\H_0=6.5\ ft$

substitute the given values

$H(t) = - \frac{1}{2}(32) t^2 + (21)t + 6.5$

$H(t) = -16t^2 + 21t + 6.5$

Rewrite

$H(t) = 6.5+21t-16t^2$