A toy rocket is launched from the top of a building 55 feet tall at an initial velocity of 242 feet per second.

Question

A toy rocket is launched from the top of a building 55 feet tall at an initial velocity of 242 feet per second.

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 654 feet above ground level?

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

Answer 1

(a)

 

h(t) = h0 + v0*t + (1/2)*a*t^2

 

h0 = 55 ft, v0 = 242 ft/s, a = -32 ft/s^2

 

h(t) = 55 + 242*t - 16*t^2 feet, with t in seconds

 

(b)

 

dh/dt = 0 = 242 - 32*t ⇒ t = 7.56 s, so maximum height is reached 7.56 s after launch 

 

hmax = h(7.56) = 55 + 242*7.56 - 16*(7.56)^2 = 970.1 ft

 

(c)

 

654 = 55 +242*t - 16*t^2

 

-16*t^2 + 242*t - 599 = 0

 

t^2 - 15.125*t + 37.438 = 0

 

t = (1/2){15.125 ± sqrt[(15.125)^2 - 4(37.438)]} = (1/2)(15.125 ± 8.889)

 

t1 = 3.118 s, t2 = 12.007 s

 

The rocket will be above 654 ft between 3.118 and 12.007 s.

 

(d)

 

0 = 55 + 242*t -16*t^2

 

t^2 - 15.125*t - 3.438 = 0

 

t = (1/2){15.125 ± sqrt[(15.125)^2 -4(-3.438)]} = (1/2)(15.125 ± 15.573)

 

Only the positive square root is a physical solution, so t = 15.349 s is when it hits the ground.