Question

a 55kg 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

Answer 1

Using Energy conservation:

KEi + PEi = KEf + PEf

KE = kinetic energy = 0.5*m*V^2

PE = potential energy = m*g*h

PEf = 0, at reference point (ground)

So,

0.5*m*Vi^2 + m*g*h = 0.5*m*Vf^2 + 0

Vi = initial speed = 72 km/hr = 72*(5/18) = 20 m/sec

Vf = final landing speed = ?

h = initial height = 8.3 m

m = mass of motorcyclist = 55 kg

So,

0.5*m*Vi^2 + m*g*h = 0.5*m*Vf^2

Vf^2 = Vi^2 + 2*g*h

Vf = sqrt (Vi^2 + 2*g*h)

Vf = sqrt (20^2 + 2*9.81*8.3)

Vf = 23.72 m/sec = 23.72*(18/5) = 85.4 km/hr

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