Question

A 1000-kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16-s, then the motor stops. The rocket altitude 20-s after launch is 510-m. You can ignore any effects of air resistance. a. What was the rocket’s acceleration during the first 16-s? b. What is the rocket’s speed as it passes through a cloud 5100-m above the ground?

Answer 1

To solve this problem we use kinematics formulas.

to. What was the acceleration of the rocket during the first 16-s?

`v_1 = v_o + a_1t_1`

Where

`v_1` = speed after 16 s

`v_0` = initial velocity = 0

`a_1` = acceleration during 16 s

`v_1 = v_0 + a_1t_1`

`v_1 = 0 + 16a_1`

`v_1 = 16a_1`

Now we use the formula for the position:

`h_1 = h_0 + v_0t_1 + 0.5a_1t_1 ^ 2`

Where:

`h_1` = position after 16 s

`h_0` = initial position = 0

`t_1` = 16 s

`h_1 = 0 + 0 + 0.5a_1(16)^2`

`h_1 = 128a_1`

Then, we know that the altitude of the rocket after 20 s is 5100 meters.

Then we will raise the equation of the position of the rocket from the instant
`t_1` to
`t_2`

`t = t_2 - t_1\\t = 20 - 16\\t = 4\ s`

where:

`h_2 = h_1 + v_1t - 0.5gt^2`

`h_2 = 128a_1 + 16a_1t - 0.5(9.8)t^2`

`5100 = 128a_1 + 16a_1(4) -0.5(9.8)(4)^2`

Now we clear
`a_1`.

`5100 +78.4 = a_1(128 + 64)`

`a_1 = 26.97 m/s^2`

The aceleration is
`26.97 m/s^2`

So:

`v_1 = a_1t\\\\v_1 = 16(26.97)\\\\v_1 = 431.52 m/s`

What is the speed of the rocket when it passes through a cloud at 5100 m above the ground?

`v_2 = v_1 - gt\\\\v_2 = 431.52 - 9.8(4)\\\\v_2 = 392.32 m/s`

The speed is 392.32 m/s