Answer to Question #103743 in Mechanics | Relativity for Rethabile Force

Question

Calculated using the value
g = 9.8 m/s2
,the time interval required for an object released from rest at some arbitrary point A,above the ground to reach the ground is Δt.

1.If g were 5.9 m/s2 , by what factor would the time interval Δt′
calculated for this value of g
differ from the value calculated using g
g = 9.8 m/s2

Answer 1

Free fall is a type of uniformly accelerated motion. The formula for the distance traveled by the body in free fall released from rest can be written as

(1) $h=\frac{g\cdot (\Delta t)^2}{2}$ .

Let point A be at the height h from the ground. Then we can solve the equation (1) regarding $\Delta t$.

(2) $\Delta t=\sqrt{\frac{2h}{g}}$

If free fall occurs on any other planet with $g=g_{'}$ we have for time interval

(3)$\Delta t^{'}=\sqrt{\frac{2h}{g_{'}}}$

Thus the time interval Δt′ differ from the value $\Delta t$ calculated using g by factor

$\frac {\Delta t^{'}}{\Delta t}=\sqrt{\frac{g}{g^{'}}}=\sqrt{\frac{9.8}{5.9}}=1.3$

Answer: The time interval Δt′ for value of $g^{'}=5.9 ms^{-2}$ differ from the value $\Delta t$ calculated using $g=9.8ms^{-2}$ by the factor 1.3. The time of free fall increased as the acceleration decrease.