Question

An object thrown upward from a height of $h$ feet with an initial velocity of $v$ feet per second will reach a maximum height of $h+\frac{v^2}{64}$ feet.

If the object is thrown upward from a height of $6$ feet with an initial velocity of $32$ feet per second, what will be its maximum height, in feet?

• A. $22$
• B. $23$
• C. $24$
• D. $25$

Answer 1

Answer: (A)

$22$

As equation is directly given, we just need to put the values directly
$h=6,v=32$
We get the maximum height
$=h+\frac{v^2}{64}$

$=6+\frac{{32}^2}{64}$

$=6+\frac{32}{2}$
$=6+16$

$=22$