A toy rocket is launched from the top of a building 64 feet tall at an initial velocity of 207 feet per second.

Question

A toy rocket is launched from the top of a building 64 feet tall at an initial velocity of 207 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

204204

feet above ground​ level?

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

Answer 1

part A:

H(t) is the height position function

V(t) = H'(t) is the velocity function

A(t) = V'(t) = H''(t) is the acceleration function

Acceleration due to gravity in feet per second squared is 32 and it decelerates.

So V'(t) = A(t) = -32

integrating:

V(t) = -32t + k for some constant k

The initial velocity is 207 , so 207 = V(0) = -32(0) + k

k = 207

H'(t) = V(t) = -32t + 207

integrating again:

H(t) = -16t^2 + 207t + k

initial height is 64 when t=0, so H(t) = -16t^2 + 207t + 64

part B:

Tmax = -b/2a = -207 / (2 * -16) = -207/-32 = 6.46875

the max height is -16*6.45875^2 + 207*6.46875 + 64 = 735.584025

part C:

-16t^2 + 207t + 64 >204

-16t^2 + 297t - 140 >0

0.716 < t < 12 and 2/9

part D:

-16t^2 + 207t + 64 = 0

t = 13.24 seconds per desmos

use the quadratic formula and a calculator