Question

modeling with polynomials quick check
a ball is thrown directly upward from the ground with an initial velocity of 4.8 ft/sec. represent the height of the ball from the ground ( t ) seconds after it was thrown upward using the model ( h(t) ). (1 point)
options:
( h(t) = -16t^2 + 4.8t )
( h(t) = -\frac{1}{2} cdot 4.8t^2 + 4.8t )
( h(t) = -\frac{1}{2} cdot 32t^2 + 4.8t + 4.8 )
( h(t) = -\frac{1}{2} cdot 0.8t^2 + 4.8t )

Explanation:

Step1: Recall the projectile motion height formula

The general formula for the height \( h(t) \) of an object in vertical motion (near the Earth's surface) is \( h(t)=-\frac{1}{2}gt^{2}+v_{0}t + h_{0} \), where \( g \) is the acceleration due to gravity, \( v_{0} \) is the initial velocity, and \( h_{0} \) is the initial height.

Step2: Identify given values

The ball is thrown from the ground, so \( h_{0} = 0 \) (initial height is 0). The initial velocity \( v_{0}=4.8\space\text{ft/sec} \). The acceleration due to gravity \( g = 32\space\text{ft/s}^2 \) (since we are using feet, the standard gravitational acceleration is \( 32\space\text{ft/s}^2 \)). Substituting into the formula:
\( h(t)=-\frac{1}{2}\times32t^{2}+4.8t + 0 \)
Simplify \( -\frac{1}{2}\times32=- 16 \), so \( h(t)=-16t^{2}+4.8t \)

Answer:

\( h(t)=-16t^{2}+4.8t \) (corresponding to the first option)