A toy rocket is launched from the top of a building 128 feet tall at an initial velocity of 205 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

160 feet above ground​ level?

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

Answer 1

Assuming that the angle of launch is 90 degrees from the ground.

The height of the rocket is given by the equation

h(t) = h(0) + V(0)t - 1/2 at² (Equation 1)

Where h(t) = the height of the rocket at time t

h(0) = the initial height of the rocket = 128 feet

V(0) = the initial velocity of the rocket = 205 ft/sec

a = acceleration due to gravity = -32.2 ft/sec²

t = time since the rocket launch in seconds

So

h(t) = 128 ft + 205 ft/sec * t - 1/2 * 32.2 ft/sec² * t² = 128 ft + 205 ft/sec * t - 16.1 ft/sec² * t²

The velocity at the peak height will be 0 feet per second.

The velocity is given by

V(t)= V(0) + a * t. (Equation 2)

Where

V(t) = The velocity at time t

V(0) is the initial velocity = 250 ft/sec

a = acceleration due to gravity = -32.2 ft/sec²

t = time in seconds

The velocity at the peak height = 0 ft/sec

Substitute known values into equation 2 and solve for t.

0 = 205 - 32.2 t

32.2 t = 205

t = 205/32.2 = 6.37 sec

Substitute t = 6.37 sec into equation 1 to find the max height

h(6.37) = 170 ft + 205 ft/s * 6.37 sec - 16.1*(6.37)² = 822.56 ft

c. Again using equation 1 we will solve for t when h(t) = 160 feet

h(t) = h(0) + v(0) * t + 1/2 * a * t²

where

h(t) = 160 ft

v(0) = 205 ft/sec

a = -32.2 ft/sec²

h(0) = 170 ft

Substituting and solving for t

160 = 170 + 205 * t - 16.1 * t²

Solve using the quadratic equation

Simplifying and putting in standard for for quadratic equation

16.1 * t² - 205 * t -10 = 0

t =( -b +/- sqt (b² - 4 * a * c))/ (2 * a)

where

sqt = square root

a = 16.1

b = -205

c = -10

t = (205 +/- sqt(205² - 4 * 16.1 * -10))/(2 * 16.1) = (205 +/- 206.56)/32.2

we can ignore the result for t < 0

t = 12.78 sec

d. Again we will use equation 1 to solve for t. h(t) in this case will be 0

0 = 170 + 205 * t - 16.1 * t² --> 16.1 * t² - 205 * t - 170 = 0

a = 16.1

b = -205

c = -170

Applying the quadratic equation we get

t = (205 +/- sqt(205² - 4 * 16.1 * -170))/(2 * 16.1) = (205 +/- 230.16)/32.2

again we do not consider t<0

t = 13.51 sec