math college algebra

Question

The Problem: 1. A toy rocket is launched from the top of a building 75 feet tall at an initial velocity of 160 feet per second. The position at time t is given by 𝑠(𝑡) = −16𝑡 2 + 𝑣0𝑡 + 𝑠0 and the velocity at time t is given by 𝑣(𝑡) = −32𝑡 + 𝑣0 where 𝑣0 is the initial velocity and 𝑠0 is the initial height. Using the position function:

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 250 feet above ground d) After how many seconds will it hit the ground? e) Sketch a graph of the path of the rocket from launch to landing. How does the graph of the path relate to the velocity of the rocket?

Next, using the velocity function: a) Give the function that describes the velocity of the rocket at time t. b) Find when the velocity is 0. c) What is the height when the velocity is 0. Does this have any significance? d) Suppose v(t) = -v0. Solve for t in this case. In particular, find t when v0 = 160 . Find s at this time. Does this value of t have significance? e) What is the velocity when the object hits the ground? f) Sketch a graph of the velocity from launch to landing. How does the graph of the velocity relate to the path of the rocket?

Answer 1

h(t)=-16t^2 +160t +75

h'(t)=v(t) = -32t+160=0

t= 160/32=5 seconds for max height

h(5)=4075 ft = max height

velocity = 0 at max height, for an instant in time, at 5 seconds

it hits the ground slightly more than 10 seconds, over double the time for max height

its velocity when it hits the ground is a little more than 160 ft/sec or a little less than -160 ft/sec

negative sign to show it's going down

h(t) = -16t^2 +160 +75 = 0 you get 2 solutions, but ignore the negative one

it's at 250 feet or over when

h(t) = -16t^2 +160t+75= 250

16t^2 -160t + 175 = 0

use the quadratic formula

t = 160/32 + or - (1/32)sqr(160^2 - 64(175))

find the 2 solutions, subtract them to get the interval where h= or > 250 feet

the graph is a downward opening parabola