Question

1) Enrique throws a stone horizontally from a cliff that has a vertical distance of 30.0m from where Enrique releases the stone to the base. The initial horizontal velocity of the stone is 6.00m/s. Apply concepts of Projectile Motion to determine

a) the horizontal distance, from the base of the cliff, the stone strikes the ground,

b) the velocity of the stone after it has fallen 10.0m. Apply the Conservation of Mechanical Energy to determine

c) the magnitude of the velocity of the stone after it has fallen 10.0m and show that it is the same value as the magnitude of the velocity that is found in part (b)

Answer 1

The time taken by the stone to reach the ground=√(2h/g)

=√(2×30/10)

=√6sec=2•45sec

Horizontal distnace covered within that time=vt=6×2•4514•7m

Velocity after it had fallen 10m

Vetical component v^2=u^2+2as

v^2=2×10×10

v=√200=14•14m/s

Horizontal component=6m/s

Resultant velocity=√200+36=√236=15•36m/s

By applying conservation of mechanical energy

At the top of cliff, potential energy=mgh=m×10×30=300m

Kinetic energy=1/2mv^2=1/2×m×36=18m

Total energy=318m

When it fallen a distance of 10meter,,potential energy=mgh=m×10×20=200m

Kinetic energy=1/2mv^2

Total enrgy=1/2mv^2+200m

Equating the mechanical energies at both point

318m=200m+1/2mv^2

118m=1/2mv^2

v=√236=15•36m/s