Question

A small sphere of mass m is launched horizontally over a body of water from a height h above the water and with a launch speed v0. Determine expressions for the following in terms of m, v0, h, and g. Air resistance is negligibly small.

(a) W is the amount work done by the force of gravity on the projectile during its flight.
(b) ΔKE is the change in kinetic energy ΔKE of the projectile from the time it was fired until it hits the water.
(c) KEf is the final kinetic energy KEf of the projectile as it hits the water.

Answer 1

Answer:

(a)

F=mgh

(b)

$\triangle KE=0.5m(v_f^{2}-v_o^{2})$

(c)

$KE_f=0.5mv_f^{2}$

Explanation:

(a)

Work done by gravity is given as the product of force and perpendicular distance hence

W=Fh

Where F is force and h is perpendicular height.

Since force due to gravity is a product of mass and acceleration due to gravity, ie F=gm then substituting this back into the original formula, we obtain

F=mgh

(b)

Kinetic energy, KE is given by

$0.5mv^{2}$

where m is the mass of an object and v is the velocity. Taking final velocitu as vf while initial velocity as given vo then change in kinetic energy is given by the difference between final and initial kinetic energy

Intial kinetic energy is $KE_o=0.5mv_o^{2}$

Final kinetic energy is

$KE_f=0.5mv_f^{2}$

Change in kinetic energy is given as

$\triangle KE=KE_f-KE_o$

$\triangle KE=0.5m(v_f^{2}-v_o^{2})$

(c)

As already derived in part b above, the final kinetic energy is equivalent to Final kinetic energy is $KE_f=0.5mv_f^{2}$