two children are throwing a baseball back and forth. the ball is 4 ft above the ground when it leaves one…

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?\nh(t) = at² + vt + h₀\n12 ft\n20 ft\n76 ft\n116 ft

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?\nh(t) = at² + vt + h₀\n12 ft\n20 ft\n76 ft\n116 ft

Answer

Explanation:

Step1: Identify the values

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

Step2: Substitute into the formula

$h(2)=-16\times2^{2}+36\times2 + 4$. First calculate $-16\times2^{2}=-16\times4=-64$. Then calculate $36\times2 = 72$. So $h(2)=-64 + 72+4$.

Step3: Simplify the expression

$h(2)=(-64 + 72)+4=8 + 4=12$.

Answer:

A. 12 ft