Question

A model rocket rises with constant acceleration to a height of 3.3m , at which point its speed is 28.0m/s .
a) What was the magnitude of the rocket's acceleration?
b) Find the height of the rocket 0.10s after launch
c) Find the speed of the rocket 0.10s after launch.

Answer 1

Question 32258

Since the rocket rises with constant acceleration, directed upwards, the velocity at given moment of time is $v(t) = at$ .

a) The y-coordinate of the rocket as a function of time is $y(t) = \frac{at^2}{2}$ . Hence, for given $y$ , time to move to it is $t_1 = \sqrt{2\frac{y}{a}}$ . Thus, according to first formula, at that moment of time velocity is $v = at_1 = a\sqrt{\frac{2h}{a}} = \sqrt{2ha}$ . This relation connects velocity with height and acceleration. Knowing that for $h = 3.3m$ velocity is $v = 28\frac{m}{s}$ , obtain $a = \frac{v^2}{2h} = 118.8\frac{m}{s^2}$ .

b) Since $y(t) = \frac{at^2}{2}$ , for $t = 0.1s$ , $y = 118.8 \cdot \frac{(0.1)^2}{2} = 0.594m$ .

c) Knowing that $v(t) = at$ , for $t = 0.1s$ , $v = 118.8 \cdot 0.1 = 11.88 \frac{m}{s}$ .