Question

A Certain rocket, initially at rest, is shot straight up with an acceleration of 6t m/s2 during the first seconds after blast-off, after which the engine cuts out and is subject

only to gravitational acceleration. How high will the rocket go? Use g=10 m/s2

Answer 1

$V = u+at = 0+6t^2$

$V^2-u^2 = 2as$

$s = \frac{6t^2}{12t} =0.5t m$

The high will the rocket go = 0.5t meter.