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A 55 kg motorcyclist is flying through the air at 72 km/h at the apex of the jump 8.3 m above the ground. Find his landing speed assuming energy is conserved.

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Answer: 85.4 km/h

Step-by-step explanation:

We would apply the law of conservation of kinetic energy. The formula is expressed as

KE1 + PE1 = KE2 + PE2

where

KE1 and KE2 are the initial and final kinetic energies

PE1 and PE2 are the initial and final potential energies

The formula for calculating kinetic energy is

KE = 1/2mv^2

where

m is the mass of the object

v is the velocity

The formula for calculating potential energy is

PE = mgh

where

g is the acceleration due to gravity = 9.8m/s^2

h is the height of the object

From the information given,

m = 55

h = 8.3

Potential energy at the top = 55 x 9.8 x 8.3 = 4473.7

At the top, v = 72km/h

Recall, 1 km = 1000m

1 hour = 3600s

72km/h = 72 x 1000/3600 = 20m/s

kinetic energy at the top = 1/2 x 55 x 20^2 = 11000

KE1 + PE1 = 4473.7 + 11000 = 15473.7

At the point of landing,

PE2 =0

KE = 1/2 x 55 x v^2 = 27.5v^2

Thus,

27.5v^2 = 15473.7

v^2 = 15473.7/27.5

v = √(15473.7/27.5)

v = 23.72 m/s

his landing speed is 23.72 m/s

Converting to km/h, we would multiply 23.72 by 18/5

Landing speed = 85.4 km/h

User Rishikesh Raje
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