Question

if a football is kicked with an initial velocity of 40 ft/s and reaches its maximum height after 1.25 seconds, what is the value of h in the equation y(t) = -16t² + vt + h? a. 16 b. 40 c. 2 d. 0

Explanation:

Step1: Find time at max height

For $y(t)=-16t^2 + vt + h$, max height at $t=-\frac{v}{2a}$. Here $a=-16$, $v=40$:
$t=-\frac{40}{2(-16)}=\frac{40}{32}=1.25$ s (matches given time).

Step2: Calculate max height

Substitute $t=1.25$, $v=40$ into $y(t)$:
$y(1.25)=-16(1.25)^2 + 40(1.25) + h$
$1.25^2=1.5625$, so $-16(1.5625)=-25$; $40(1.25)=50$
$y(1.25)=-25 + 50 + h=25 + h$

Wait—correction: The problem says "what is the value of h", but actually, since the max height time is given as 1.25s which is exactly the vertex time from $v=40$, but wait no—wait, no, the question must be asking for max height? Wait no, the options: a.16, b.40, c.2, d.0. Wait no, wait the vertex time is $t=v/(32)=40/32=1.25$, which is given, so the max height is $y(1.25)=-16*(1.25)^2 +40*(1.25) +h$. But if h=0 (initial height), then $y= -25 +50=25$, not in options. Wait wait—maybe the question was mistyped? Wait no, wait the original question: "what is the value of h"—but if initial height h=0, then option d.0. Because when kicking a football, initial height h is usually 0 (ground level).

Answer:

d. 0