Eric C. answered • 02/24/16

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Hi Kelani.

Your formula of interest is:

h(t)= −4.9t^2 + 106t + 388

a) Initially launched is a code term for "t=0". Plug in 0 for t and determine h(t)

h(0) = -4.9(0)^2 + 106(0) + 388

h(0) = 388

b) The rocket will reach maximum height based on a formula you may or may not remember from Algebra 1 concerning quadratic equations:

t_max = -b/2a

In your case, b = 106, a = -4.9

So

t_max = -106/-9.8

t_max = 10.82 seconds

The other way to determine the max height is to take the derivative of h(t) and set it equal to 0.

h'(t) = -9.8t + 106

h'(t) = 0

0 = -9.8t + 106

9.8t = 106

t = 106/9.8

t = 10.82 seconds

So you yield the same result.

c) The maximum height will occur at the maximum time. So you want to plug in t_max for h(t)

h(t_max) = -4.9*(10.82)^2 + 106(10.82) + 388

h(t_max) = 961.7 ft

d) Determining when it splashes is the same thing as asking when the height of the rocket is 0. So set h(t) equal to 0 and determine the t's which make it happen.

0 = −4.9t^2 + 106t + 388

Here, since you have a quadratic equation, you can employ the quadratic formula to solve for t.

t = (-b +/- √(b^2 - 4ac))/2a

a = -4.9

b = 106

c = 388

t = -3.2, 24.8

Time can't be negative so we can throw the first term out.

Your rocket will splash down in 24.8 seconds.

Hope this helps.