If you fire a paper clip straight up into the air with a rubber band, its height (in feet) above your hand time t seconds is given by: h = 64t - 16t^2.


1) How long does it take the paper clip to reach its maximum height?

2) If your hand is 6 feet above the ground at "launch", what is the maximum height the paper clip attains?

3) At what 2 times is the paper clip traveling at a speed of 24 ft/sec?

4) What was the initial velocity imparted to the paper clip by the rubber band?

#1 as with any parabola, vertex is at t = -b/2a

#2 use the t from #1 and evaluate h(t)+6

#3 dh/dt = 64-32t
when is that 24 or -24?

#4 h(t) = vt-g/2 t^2

To solve these questions, we need to understand the equation h = 64t - 16t^2, which represents the height of the paper clip above your hand at time t seconds.

1) To find the time it takes for the paper clip to reach its maximum height, we need to determine the vertex of the parabolic equation. The vertex occurs at the maximum point of the parabola. The formula for finding the x-coordinate of the vertex is given by -b/2a in the equation ax^2 + bx + c = 0. In our case, the equation is -16t^2 + 64t. Comparing it with the general form, we can see that a = -16 and b = 64.

So, the t-coordinate of the vertex is -b/2a = -64/(2*(-16)) = 64/32 = 2 seconds. Therefore, it takes 2 seconds for the paper clip to reach its maximum height.

2) To find the maximum height the paper clip attains, we substitute the value t = 2 into the equation.

h = 64t - 16t^2
h = 64(2) - 16(2^2)
h = 128 - 16(4)
h = 128 - 64
h = 64 feet

Thus, the maximum height the paper clip attains is 64 feet.

3) To find the times at which the paper clip is traveling at a speed of 24 ft/sec, we need to determine when the derivative of height with respect to time (dh/dt) is equal to 24.

Taking the derivative of the equation h = 64t - 16t^2, we get dh/dt = 64 - 32t. We set this derivative equal to 24 and solve for t:

64 - 32t = 24
-32t = 24 - 64
-32t = -40
t = (-40)/(-32)
t = 1.25 seconds

So, the paper clip is traveling at a speed of 24 ft/sec at t = 1.25 seconds.

To find the second time, we have -32t = -24 (similar equation as above, but with a different value). Solving for t:

t = (-24)/(-32)
t = 0.75 seconds

Thus, the paper clip is traveling at a speed of 24 ft/sec at t = 1.25 seconds and t = 0.75 seconds.

4) The initial velocity imparted to the paper clip by the rubber band can be determined by examining the coefficient of the linear term (t term) in the equation. In our equation h = 64t - 16t^2, the coefficient of the t term is 64.

Therefore, the initial velocity imparted to the paper clip by the rubber band is 64 ft/sec.