Question

A model rocket is launched straight upward. The solid fuel propellant pushes the rocket off the ground at an initial velocity of 50 feet per second.

s(t)= -4t2 + 50t - 84. Show work to get credit


a. What is the maximum height reached by the rocket?

b. What time will it reach the maximum height?

c. You forgot to put the parachute in the rocket. When will the rocket hit the ground?

Answer 1

a) 72.25sec

b) 6.25secs

c) after 10.5secs and 2 secs

Explanation:

Given the height reached by the rocket expressed as;

s(t)= -4t^2 + 50t - 84

At maximum height, the velocity of the rocket is zero i.e ds/dt = 0

ds/dt = -8t + 50

0 = -8t + 50

8t = 50

t = 50/8

t = 6.25secs

Hence it will reach the maximum height after 6.25secs

To get the maximum height, you will substitute t - 6.25s into the given expression

s(t)= -4t^2 + 50t - 84

s(6.25) = -4(6.25)^2 + 50(6.25) - 84

s(6.25) = -156.25 + 312.5 - 84

s(6.25) = 72.25feet

Hence the maximum height reached by the rocket is 72.25feet

The rocket will reach the ground when s(t) = 0

Substitute into the expression

s(t)= -4t^2 + 50t - 84

0 = -4t^2 + 50t - 84

4t^2 - 50t + 84 = 0

2t^2 - 25t + 42 = 0

2t^2 - 4t - 21t + 42 = 0

2t(t-2)-21(t-2) = 0

(2t - 21) (t - 2) = 0

2t - 21 = 0 and t - 2 = 0

2t = 21 and t = 2

t = 10.5 and 2

Hence the time the rocket will reach the ground are after 10.5secs and 2 secs