Question

1- imagine that the car start from rest and height is 40m , calculate the speed of the car at the bottom of the hill , 2- at what height it will leave half this speed ?
show all your steps ..

Answer 1

Part A.

Using energy conservation at the top and bottom of hill

KEi + PEi = KEf + PEf

PEf = 0, at ground

KEi = 0, initially speed is zero

So

PEi = KEf

m*g*h = 0.5*m*V^2

V = sqrt (2*g*h)

given that h = 40 m

So,

V = sqrt (2*9.8*40)

V = 28 m/sec = Speed of car at the bottom of hill

Part B.

we need height 'h1' when V1 = V/2 = 14 m/sec

Again Using energy conservation,

KEi + PEi = KEf + PEf

KEi = 0, since initial speed is zero

PEi = m*g*h

PEf = m*g*h1

KEf = 0.5*m*V1^2

So,

0 + m*g*h = m*g*h1 + 0.5*m*V1^2

h1 = h - 0.5*V1^2/g

h1 = 40 - (0.5*14^2/9.8)

h1 = 30 m = height from bottom when speed is half of the final speed

h1 = 40 - 30 = 10 m = height from top of the hill

Check which height you need from the bottom or from the top

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