A truck, initially at rest, rolls down a frictionless hill and attains a speed of 2 m/s at the bottom. To achieve a speed of 6 m/s at the bottom, how many times higher must the hill be?

so the height will be increased by 9 times

Explanation:

The truck is in its pontential energy at rest

The potential energy is converted to kinetic energy when in motion

Given that,

 h = height of the hill

m = mass of the truck

g = acceleration due to gravity

v = velocity of the truck at the bottom of the hill.

At the top of the hill,

Potential energy(PE) = mgh

Kinetic energy(KE) = 0

Total energy, E = PE + KE = mgh + 0 = mgh(1)

At the bottom of the hill :

Potential energy, PE = 0  

Kinetic energy,KE = (1/2)mv²

Total energy, E = PE + KE

= 0 + (1/2)mv²(2)  

In law of conservation of energy,

(1) = (2)

 mgh = (1/2)mv²

h = (1/2g)v²

given,

v = 2 m/s

g = 9.81 m/s²

h = (1/2(9.81)) × 2²

h = 0.20 m

so,

if v = 6 m/s

h =  (1/2(9.81)) × 6²

h = 1.83 m

from the first h = 0.20m and the second h = 1.83m

1.83m / 0.20m

= 9.15

so the height will be increased by 9 times

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