Question

An archer shoots an row into the air such that it's height at any time, t, is given by the function h (t)= -16t^2+kt+3. If the maximum height of the arrow occurs at time t=4, what is the value of k?

A) 128
B) 64
C) 8
D)4

What I did:
H (t)= -16t^2+kt+3
4= -16 (4)^2+ k(4)+3
4= -16 (16)+ k(4)+3
4= -256+ k(4)+ 3
4= -253+ k(4)
0=-257+4k
257=4k
64=k

K is 64.

My answer key says 128 as the answer. When I plugged in 128 as k and graphed it on my graphing calculator.the maximum height so the vertex was (4,128). But how did I get 64 then. Please explain and show me the steps.

Thank you

Answer 1

Why are you letting the left side equal to 4,
you want to find the left side of your equation.

h(t) is in metres, and t is in time, so you mixed up the units.

Ok, so you know that the vertex is (4, ?)
The t of the vertex is found by -b/(2a)
= -k/-32

but we are told that the t of the vertex is 4
-k/-32 = 4
k/32 = 4
k = 128

Answer 2

H is the height, not t. You used t = 4 for H

another way (dH/dt) = v
velocity = v = -32t + k
at top:
v = 0 = -32 t + k
k = 32 t = 32 * 4 = 128

Answer 3

I did it the physics way.
Use Reiny's parabola if you have not had physics and calculus.