A toy rocket is launched wih an upward velocity of 112 feet

A toy rocket is launched wih an upward velocity of 112 feet per second. The height \"h\", in feet of the rocket \"t\" seconds after being launched is given by h(t)=-16t^2+112t+97.

-Rewrite h(t) in vertex form

-When does the rocket reach its maximum height and how high is the rocket then? Explain.

-Calculate the total length of the trip in seconds algebraically

Solution

h(t)=-16t^2+112t+97.

h(t) = -16(t^2 - 7t) +97

=-16(t^2 -7t +49/4 -49/4) +97

= -16(t^2 -7t+49/4) +196 +97

h(t) =-16( t- 7/2)^2 +293

It is as per the standard vertex form of equation: y = a(x-h)^2 +k

--- rocket reach its maximum height at point corresponding to vertex t = 7/2 sec

maximum height H = 293 feet

----total length of the trip is completed when h(t) =0

0 =-16( t- 7/2)^2 +293

18.31 = (t -7/2)^2

4.28 = t -3.5

t = 7.78 seconds total length of the trip