The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '. D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .--
They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.
Answer:
Explanation:
The minimum magnitude of acceleration = 3 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 3 x 1²
= - 3 + 1.5
= - 1.5 m
position at t = 1 s
= 10 - 1.5
= 8.5 m
The maximum magnitude of acceleration = 6 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 6 x 1²
= - 3 + 3
= 0
position at t = 1 s
= 10 +0
= 10 m
So range of position is 8.5 m to 10 m .
That particular strike was very roughly 2.4 km (1.5 miles) away from them.
That's if you use 340 m/s (1120 ft/sec) for the speed of sound.
But the air in the region for several thousand feet around a thunderstorm
is doing weird things to sounds that pass through it, so you can't use any
exact number for the speed of sound in a stormy area.
The only thing you can be absolutely sure of is that Johnny and his friends
need to round up their equipment and get in the house. NOW !
