A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 60t at which time the fuel is exhausted and it becomes a freely "falling" body.

Question

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 60t at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to -18 ft/s in 5 s. The rocket then "floats" to the ground at that rate.

(a) At what time does the rocket reach its maximum height? (Give your answer correct to one decimal place.)

 

(b)What is that height? (Give your answer correct to the nearest whole number.)

 ft

(c) At what time does the rocket land? (Give your answer correct to one decimal place.)

Answer 1

UPDATED:

Question:

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 60t at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to -18 ft/s in 5 s. The rocket then "floats" to the ground at that rate.

Solution:

We must break this scenario up into 5 stages.

I like to follow the GUESS Method: Givens, Unknowns, Equations, Solve, Substitute.

GIVENS:

Stage 1: Rocket launches with thrusters

t = 3 sec

a(t) = 60t

v(0) = 0 ft/s (rest)

y(0) = 0 ft (rest)

Stage 2: Rocket still climbing, but no thrusters (begin free fall)

t = ? sec

Stage 3: Rocket reached highest altitude, now falling downward

t = 14 sec (free fall)

a(t) = g = 32.2 ft/s^2 (free falling)

Stage 4: Parachute Opens (Linear Slowdown)

t = 5 sec

v(t) = From v_stage3 down to -18 ft/s

Stage 5: Velocity slows down to a constant until it lands

t = ? sec

v(t) = -18 ft/sec

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(a) At what time does the rocket reach its maximum height? (Give your answer correct to one decimal place.)

UNKNOWNS: t_y_max

EQUATIONS:

y(t) = y(0) + v(t)*t + (1/2)*a*t^2

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Stage 1: Rocket launches with thrusters

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SOLVE:

a(t) = Rocket Up - gravity down

y(t) = y(0) + v(t)*t + (1/2)*a*t^2

y(t) = 0 + 0*t + (1/2)*(60t - 32.2)*t^2

y(t) = (30*t - 16.1)*t^2

y(t) = 30*t^3 - 16.1*t^2

SUBSTITUTE:

t_stage1 = 3 sec

y(3) = 30*3^3 - 16.1*3^2

y(3) = 665.1 ft high

v(t) = y'(t) = 90*t^2 - 32.2*t

At t = 3 seconds:

v(3) = 90*3^2 - 32.2*3

= 90*9 - 96.6

= 810.0 - 96.6

v(3) = 713.4 ft/sec upwards

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Stage 2: Rocket still climbing, but no thrusters (begin free fall)

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SOLVE:

Max height is when v(t) = 0.

a(t) = gravity down

y(t) = y(0) + v(t)*t - (1/2)*g*t^2

SUBSTITUTE:

y(t) = 665.1 + 713.4*t - (1/2)*(32.2)*t^2

y(t) = 665.1 + 713.4*t - 16.1*t^2

v(t) = y'(t) = 713.4 - 32.2*t

v(t) = 713.4 - 32.2*t = 0

= 713.4/32.2

t_stage2 = 22.155 sec when it stops moving upwards

Total time climbing = t_stage + t_stage2

= 3 sec + 22.155 sec

= 25.155 sec

THEREFORE, the rocket reach its maximum height at 25.2 seconds

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(b) What is that height? (Give your answer correct to the nearest whole number.) in ft

UNKNOWNS: y_max

y_stage1 = y(3) = 665.1 ft

y_stage2 = y(t) = 665.1 + 713.4*(t_stage2) - 16.1*(t_stage2)^2

= 665.1 + 713.4*(22.155) - 16.1*(22.155)^2

= 8,567.69

y_stage2 = 8,567.7 ft

y_max = y_stage1 + y_stage2

= 665.1 ft + 8,567.7 ft

= 9,232.79 ft

THEREFORE, the maximum rocket height is 9,233 feet high

=================================================================================

(c) At what time does the rocket land? (Give your answer correct to one decimal place.)

UNKNOWNS: t_total

EQUATIONS:

t_total = t_stage1 + t_stage2 + t_stage3 + t_stage4 + t_stage5

t_stage1 = 3 sec (given)

t_stage2 = 22.155 sec

t_stage3 = 14 sec (given)

t_stage4 = 5 sec

t_stage5 = ? sec

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Stage 3: Rocket reached highest altitude, now falling downward

*****************************************************************************

EQUATIONS:

a(t) = gravity downward

y(t) = y(t) + v(t)*t - (1/2)*g*t^2

y(t) = y_max = 9,232.79 ft

v(t) = 0 ft/sec (stopped climbing)

SUBSTITUTE:

y(t) = 9,232.79 ft - (1/2)*(32.2)*t^2 ft

y(t) = 9,232.79 ft - 16.1*t^2 ft

y(t) = 9,232.79 ft - 16.1*14^2 ft

t = 14 sec (Given: 14 sec free fall)

= 9,232.79 - 16.1*14^2 ft

= 9,232.79 - 3,155.6 ft

= 6,077.188 ft

In 14 seconds the rocket falls 3,155.6 ft to a height of 6,077.2 feet

v(t) = y'(t) = (- 16.1*t^2)'

= -32.2*t

v(14) = -32.2*(14)

= -450.8 ft/sec

The velocity 14 seconds after the rocket reaches its highest altitude is -450.8 ft/sec (downward)

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Stage 4: Parachute Opens (Linear Slowdown)

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v(t) slows to -18 ft/s in 5 sec

t = 5 sec

EQUATIONS:

Linear Equation for Velocity:

v(0) = -450.8 ft/sec

v(5) = -18 ft/sec

velocity slope = [v(5) - v(0)] / 5 sec

= [-18 -(-450.8)] / 5

= [-18 + 450.8] / 5

= 432.8 / 5

= 86.56 ft/sec

v(t) = -450.8 + 86.56*t ft/sec

y(t) = 6,077.188 + v(t)*t - (1/2)*g*t^2 ft

SUBSTITUTE:

y(t) = 6,077.188 + (-450.8 + 86.56*t)*t - 16.1*t^2 ft

y(t) = 6,077.188 - 450.8*t + 86.56*t^2 - 16.1*t^2 ft

y(t) = 6,077.188 - 450.8*t + 70.48*t^2 ft

t = 5 sec

y(t) = 6,077.188 - 450.8*5 + 70.48*5^2 ft

y(t) = 5,584.688 ft

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Stage 5: Velocity slows down to a constant until it lands

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a(t) = 0 (floats down at a constant velocity since parachute cancels gravity)

v(t) = -18*t

y(t) = 5,584.7 ft

y(t) = y(0) + v(t)*t + (1/2)*a*t^2

y(t) = 5,584.7 - 18*t - (1/2)*0*t^2 ft

y(t) = 5,584.7 - 18*t

When y(t) = 0 it has landed

5,584.7 - 18*t = 0

t = 5,584.7 / 18

t = 310.3 sec

t_total = t_stage1 + t_stage2 + t_stage3 + t_stage4 + t_stage5

t_stage1 = 3 sec (given)

t_stage2 = 22.155 sec

t_stage3 = 14 sec (given)

t_stage4 = 5 sec

t_stage5 = 310.3 sec

t_total = 3 + 22.2 + 14 + 5 + 310.3

= 354.5 sec

THEREFORE, the rocket lands 354.5 seconds after launch (just under 6 minutes)