A model rocket is launched with an initial upward velocity of 195 ft/s. The rocket's height h (in feet) after t seconds is given
by the following
h= 1951 - 1612
Find all values of 1 for which the rocket's height is 87 feet.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
1 =
seconds
x
5
?
ground

  0.46 seconds or 11.72 seconds

Step-by-step explanation:

Putting 87 for h in the height equation gives ...

  87 = -16t^2 +195t

Dividing by -16 puts this in a little nicer form for finding solutions.

  t^2 -(195/16)t = -87/16

To "complete the square", we can add the square of half the coefficient of t:

  t^2 -(195/16)t + (195/32)^2 = (195/32)^2 -87/16

  (t -195/32)^2 = 32457/1024

Taking the square root, we have ...

  t -195/32 = ±(1/32)√32457

  t = (195 ±√32457)/32 ≈ 0.4638 or 11.7237

The rocket is at a height of 87 feet after 0.46 seconds, and again after 11.72 seconds.

rx and sx

step-by-step explanation:

your answer would be 12