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
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
Answers
Reiny
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
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
Damon
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
another way (dH/dt) = v
velocity = v = -32t + k
at top:
v = 0 = -32 t + k
k = 32 t = 32 * 4 = 128
Damon
I did it the physics way.
Use Reiny's parabola if you have not had physics and calculus.
Use Reiny's parabola if you have not had physics and calculus.