Question

A model rocket is launched from a platform 12 meters high at a speed of 35 meters per second. Its height h can be modeled by the equation h= -4.9t'2+35t+12 where t is the time in seconds. at what time will the rocket be altitude of 60 meters.

Answer 1

Answer:

Step-by-step explanation:

so set the given equation equal to 60 and solve for t

60= -4.9$t^{2}$ +35t +12

0= -4.9$t^{2}$ +35t - 48

now use the quadratic equation to solve for t

-b +- sq rt ( $b^{2}$ - 4 a c )  /  2 a

t = -35 +- sq. rt [ 1225 -4(-4.9)(-48) ] / 2 (-4.9)

t = -35 +- sq. rt [ 284.2] / -9.8

t = -35 +- 16/858 / - 9.8

t = 3.571 +- 1.72

t  = 1.85 seconds ( on the way up ) or  t = 5.29 seconds ( on the way down)

at both of those times the rocket will be at 60 meters  :)