Assuming no air resistance, how far will a 0.0010 kg raindrop have fallen when it hits the ground 30.0 s later. assume g = 9.8 m/s2.

a)294 m
b)4410 m
c)8820 m
d)44,000 km

Option b, 4410 meters.

Explanation: If there is no air resistance, the acceleration experienced by the raindrop is equal to the gravitational acceleration:

a = -9.8m/s^2

for the velocity, we integrate over time, and there is no constant because we don't have initial velocity.

v = - (9.8m/s^2)*t

For the position, we integrate again over time, and we add a constant of integration that will be equal to the initial height of the raindrop.

r = (-9.8m/s^2)(t^2)/2 + H

if at t= 30s the raindrop reaches the ground, we have that:

r(30) = 0 = (-4.9*900)m + H

H = 4.9*900m = 4410m

So the initial height of the raindrop s 4410 meters.

a and b sometimes have a positive value

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.