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 of variables

Given $a=-16$, $v = 36$, $h_0=4$, $t = 2$.

Step2: Substitute values into the formula

$h(t)=at^{2}+vt + h_0$ becomes $h(2)=-16\times2^{2}+36\times2 + 4$.

Step3: Calculate each term

First term: $-16\times2^{2}=-16\times4=-64$. Second - term: $36\times2 = 72$.

Step4: Find the value of $h(2)$

$h(2)=-64 + 72+4=12$.

Answer:

12 ft