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