Question

A feather is dropped on the planet other than earth which has very low acceleration due to gravity from a height of 1.40 meters. The acceleration of gravity on the other planet is 1.67 m/s2. Determine the time of feather to fall to the surface of the other planet

Answer 1

s = ut + 1/2 at^2
where s is the distance, u is the initial velocity, t is time and a is acceleration

so we have
u = 0
so

$1.4 = 0t + \frac{1}{2} \times 1.67 \times {t}^{2} \\ 1.4 = 0 + \frac{1}{2} \times 1.67 \times {t}^{2} \\ 1.4 = 0 + 0.835 {t}^{2} \\ 0.835 {t}^{2} = 1.4 \\ {t}^{2} = 1.4 \div 0.835 = 1.67 \\ t = \sqrt{1.67} = 1.29$
time is 1.29 = 1.3
1.3 seconds approx time

Hope it helps

Answer 2

Hi there
check it out....