Question

two children are throwing a baseball back and forth. the ball is 4 ft above the ground when it leaves one childs hand with an upward velocity of 36 ft/s. if acceleration due to gravity is -16 ft/s², how high above the ground is the ball 2 s after it is thrown?

$h(t) = at² + vt + h₀$

○ 12 ft
○ 20 ft
○ 76 ft
○ 116 ft

Explanation:

Step1: Identify the values

We know the formula for the height of the ball is \( h(t) = at^2 + vt + h_0 \), where \( a \) is the acceleration, \( v \) is the initial velocity, \( h_0 \) is the initial height, and \( t \) is the time. Here, \( a=- 16\space ft/s^{2}\), \( v = 36\space ft/s\), \( h_0=4\space ft\) and \( t = 2\space s\).

Step2: Substitute the values into the formula

Substitute \( a=-16\), \( v = 36\), \( h_0 = 4\) and \( t=2\) into the formula \( h(t)=at^{2}+vt + h_0\).

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Answer:

12 ft