Question

modeling with polynomials quick check
item 1 (1 point)
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) )
options:
( h(t) = -\frac{1}{2} cdot 4.8t^2 + 4.8t )
( h(t) = -16t^2 + 4.8t )
( h(t) = -\frac{1}{2} cdot 9.8t^2 + 4.8t )
( h(t) = -\frac{1}{2} cdot 32t^2 + 4.8t + 4.8 )

Explanation:

Step1: Recall the projectile motion 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.
For objects near the Earth's surface, the acceleration due to gravity \( g = 32\space\text{ft/s}^2 \) (when using feet as the unit of length). The ball is thrown from the ground, so the initial height \( h_{0}=0 \), and the initial velocity \( v_{0} = 4.8\space\text{ft/s} \).
Substituting \( g = 32 \), \( v_{0}=4.8 \) and \( h_{0} = 0 \) into the formula, we get:
\( h(t)=-\frac{1}{2}\times32t^{2}+4.8t+0 \)
Simplify \( -\frac{1}{2}\times32 \), we have \( - 16 \), so \( h(t)=-16t^{2}+4.8t \)

Step2: Match with the given options

Looking at the options, the function \( h(t)=-16t^{2}+4.8t \) is one of the options (the third option as per the image).

Answer:

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