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A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16.5 s later. The descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude. Assume that g = 32.2 ft/s2.

Determine
(a) the speed v1 of the rocket at the end of powered flight,
(b) the maximum altitude reached by the rocket.

User Ravindu
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1 Answer

1 vote

Answer:


u = 260.22m/s


S_(max) = 1141.07ft

Step-by-step explanation:

Given


S_0 = 89.6ft --- Initial altitude


S_(16.5) = 0ft -- Altitude after 16.5 seconds


a = -g = -32.2ft/s^2 --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:


S = ut + (1)/(2)at^2

The final altitude after 16.5 seconds is represented as:


S_(16.5) = S_0 + ut + (1)/(2)at^2

Substitute the following values:


S_0 = 89.6ft
S_(16.5) = 0ft
a = -g = -32.2ft/s^2 and
t = 16.5

So, we have:


0 = 89.6 + u * 16.5 - (1)/(2) * 32.2 * 16.5^2


0 = 89.6 + u * 16.5 - (1)/(2) * 8766.45


0 = 89.6 + 16.5u- 4383.225

Collect Like Terms


16.5u = -89.6 +4383.225


16.5u = 4293.625

Make u the subject


u = (4293.625)/(16.5)


u = 260.21969697


u = 260.22m/s

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:


v=u + at

At the maximum height:


v =0 --- The final velocity


u = 260.22m/s


a = -g = -32.2ft/s^2

So, we have:


0 = 260.22 - 32.2t

Collect Like Terms


32.2t = 260.22

Make t the subject


t = (260.22)/( 32.2)


t = 8.08s

The maximum height is then calculated as:


S_(max) = S_0 + ut + (1)/(2)at^2

This gives:


S_(max) = 89.6 + 260.22 * 8.08 - (1)/(2) * 32.2 * 8.08^2


S_(max) = 89.6 + 260.22 * 8.08 - (1)/(2) * 2102.22


S_(max) = 89.6 + 260.22 * 8.08 - 1051.11


S_(max) = 1141.0676


S_(max) = 1141.07ft

Hence, the maximum height is 1141.07ft

User Iku
by
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