What’s the velocity of a bus travelling through a town with a mass of 5040 kg and kinetic energy of 493,900 j ?

The velocity of the bus is 14 m/s

Explanation:

Kinetic energy is directly proportional to the mass of the object and to

the square of its velocity

→ K.E = mv², where K.E is the kinetic energy, m is the

mass, and v is the velocity

The unit of the kinetic energy is joule (j)

We need to find the velocity of a bus travelling through a town with a

mass of 5040 kg and kinetic energy of 493,900 j

→ K.E = 493900 j , m = 5040 kg

Substitute these values in the equation above

→ 493900 = (5040)v²

Multiply both sides by 2

→ 987800 = (5040)v²

Divide both sides by 5040

→ 196.00 = v²

Take √ for both sides

→ v = 14 m/s

The velocity of the bus is 14 m/s

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft