A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K. E = 1/2 * m * v2

Where m is mass

v is velocity

Hence.

K. E = 1/2 * 2.25 * (187.5) ^2

Now this should be same with the potential energy which is given as;

P. E = m * g * h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P. E = 2.25 * 9.8 * h

From the foregoing analysis of energy conversation it implies;

1/2 * 2.25 * (187.5) ^2 = 2.25 * 9.8 * h

=> 1/2 * (187.5) ^2 = 9.8 * h

=>1/2 * (187.5) ^2 / 9.8 = h

=> 1793.69m = h

h = 1793.69m

h = 1793.7m to 1 decimal place