To solve this problem we use kinematics formulas.
to. What was the acceleration of the rocket during the first 16-s?
[tex]v_1 = v_o + a_1t_1[/tex]
Where
[tex]v_1[/tex] = speed after 16 s
[tex]v_0[/tex] = initial velocity = 0
[tex]a_1[/tex] = acceleration during 16 s
[tex]v_1 = v_0 + a_1t_1[/tex]
[tex]v_1 = 0 + 16a_1[/tex]
[tex]v_1 = 16a_1[/tex]
Now we use the formula for the position:
[tex]h_1 = h_0 + v_0t_1 + 0.5a_1t_1 ^ 2[/tex]
Where:
[tex]h_1[/tex] = position after 16 s
[tex]h_0[/tex] = initial position = 0
[tex]t_1[/tex] = 16 s
[tex]h_1 = 0 + 0 + 0.5a_1(16)^2[/tex]
[tex]h_1 = 128a_1[/tex]
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 [tex]t_1[/tex] to [tex]t_2[/tex]
[tex]t = t_2 - t_1\\t = 20 - 16\\t = 4\ s[/tex]
where:
[tex]h_2 = h_1 + v_1t - 0.5gt^2[/tex]
[tex]h_2 = 128a_1 + 16a_1t - 0.5(9.8)t^2[/tex]
[tex]5100 = 128a_1 + 16a_1(4) -0.5(9.8)(4)^2[/tex]
Now we clear [tex]a_1[/tex].
[tex]5100 +78.4 = a_1(128 + 64)[/tex]
[tex]a_1 = 26.97 m/s^2[/tex]
The aceleration is [tex]26.97 m/s^2[/tex]
So:
[tex]v_1 = a_1t\\\\v_1 = 16(26.97)\\\\v_1 = 431.52 m/s[/tex]
What is the speed of the rocket when it passes through a cloud at 5100 m above the ground?
[tex]v_2 = v_1 - gt\\\\v_2 = 431.52 - 9.8(4)\\\\v_2 = 392.32 m/s[/tex]
The speed is 392.32 m/s
