Question

Jeff is part of a trekking team. As he climbs a hill, he drops his water bottle, which has a mass of 0.25 kilograms. The bottle slides down the hill and is moving at a velocity of 14 meters/second the instant it hits the ground. The vertical height from where Jeff dropped the bottle is meters. Ignore friction, and use , PE = m × g × h, and g = 9.8 m/s2.

Answer 1

Answer:

h = 10 m

Explanation:

As per energy conservation law we know that if there is no friction force present in the system then in that case total energy is always remains conserved.

So here we can say that

initial potential energy = final kinetic energy

$\frac{1}{2}mv^2 = mgh$

here we know that

m = 0.25 kg

v = 14 m/s

now from above equation

$\frac{1}{2}(0.25)(14)^2 = 0.25(9.8)(h)$

$h = 10 m$

so initial vertical height will be h = 10 m

Answer 2

Answer:

h=10m

1/2(0.25)(14)^2=0.25(9.8)(h)