Question

A model rocket rises with constant acceleration to a height of 3.1 m, at which point its speed is 28.0 m/s. a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket's acceleration?
c. Find the height of the rocket 0.10 s after launch.
d. Find the speed of the rocket 0.10 s after launch.

Answer 1

Explanation:

It is given that,

Height, h = 3.1 m

Initial speed of the rocket, u = 0

Final speed of the rocket, v = 28 m/s

(b) Let a is the acceleration of the rocket. Using the formula as :

$a=\dfrac{v^2-u^2}{2h}$

$a=\dfrac{(28)^2}{2\times 3.1}$

$a=126.45\ m/s^2$

(a) Let t is the time taken to reach by the rocket to reach to a height of h. So,

$t=\dfrac{v-u}{a}$

$t=\dfrac{28\ m/s}{126.45\ m/s^2}$

t = 0.22 seconds

(c) At t = 0.1 seconds, height of the rocket is given by :

$h=ut+\dfrac{1}{2}at^2$

$h=\dfrac{1}{2}\times 126.45\times (0.1)^2$

h = 0.63 meters

(d) Let v' is the speed of the rocket 0.10 s after launch.

So, $v'=u+at$

$v'=0+126.45\times 0.1$

v' = 12.64 m/s

Hence, this is the required solution.