A toy rocket is launched from the top of a building 62 feet tall at an initial velocity of 245 feet per second. ​

Question

a) Give the function that describes the height of the rocket in terms of time t.

​b) Determine the time at which the rocket reaches its maximum​ height, and the maximum height in feet. ​

c) For what time interval will the rocket be more than 421 feet above ground​ level?

d) After how many seconds will it hit the​ ground?

Answer 1

H(t) = -16t^2 + 245 t + 62

THe max height is almost 1000 ft. at 7.656 seconds

B=245 A = -16 , so max occurs at t = -B/(2a) = 7.65625

-16t^2 + 245 t + 62 - 421 = 0

-16t^2 + 245 t - 359 =0

16t^2 - 245 t + 359 = 0

[245 +or- sqrt( (-245)^2 - 4(16)(359))/32

it is at or above 421 feet when

1.641213502812306293395648279763..< t < 13.671286497187693706604351720237...

Finally -16t^2 + 245 t + 62=0

when t = 15.562 per quadratic formula

Answer 2

a) Hite y(t) = 62 + 245t - 16t²; acceleration of gravity g = 32 ft/s²(origi at the grount)

b) Velocity v = 245 - 32t. At maximum height v = 0, so 245 - 32t = 0, t = 245/32 s = 7.66 s.then

maximum he h_max(7.66) = 62 + 245·7.66 - 16·7.66² = 999.9 feet.

c) y(t) = 62 + 245t - 16t² > 421; 16t² - 245t + 359 < 0.

I solve this inequality uzing TI-84 1.63 s < t < 13.67 s

d) y(t) = 62 + 245t - 16t² = 0, t = 15.56 s