Question

A toy rocket moving vertically upward passes by a 2.0-m-high window whose sill is 8.0 m above the ground. The rocket takes 0.15 s to travel the 2.0 m height of the window. What was the launch speed of the rocket, and how high will it go? Assume the propellant is burned very quickly at blastoff.

Answer 1

Answer:

$v_i = 18.86 m/s$

Explanation:

As we know that the speed of the rocket is v1 and v2 at the bottom and top of the window

then we will have

$d = (\frac{v_1 + v_2}{2}) t$

$2 = (\frac{v_1 + v_2}{2})(0.15)$

$26.67 m/s = v_1 + v_2$

also we know that

$v_2 - v_1 = (-9.81)(t)$

$v_2 - v_1 = (-9.81)(0.15) = -1.47$

now we have

$v_2 = 12.6$

also we have

$v_1 = 14.1 m/s$

now if the sill of the window is at height 8 m from the ground then we have

$v_1^2 - v_i^2 = 2 a h$

$(14.1^2) - v_i^2 = 2(-9.81)(8)$

$v_i = 18.86 m/s$