Question

In a certain instant, a mortar projectile develops a velocity of 60 m/s. If the potential energy at that point is half of its kinetic energy, what is the height above the ground?

a) 90 m
b) 120 m
c) 180 m
d) 240 m

Answer 1

The height above the ground where the potential energy is half the kinetic energy of a mortar projectile with a velocity of 60 m/s is approximately 90 meters.

Step-by-step explanation:

To calculate the height above the ground where the potential energy (PE) is half the kinetic energy (KE) of the mortar projectile, we need to use the relationship PE = KE/2 and the equation for kinetic energy KE = (1/2)mv2. Given that the velocity (v) is 60 m/s and no mass is provided, we assume a unit mass for simplicity. Therefore, KE = (1/2)(60 m/s)2 = 1800 J. Accordingly, the potential energy PE would be 900 J.

Since potential energy is also defined as PE = mgh, where m is the mass (which can be taken as 1 kg for easy calculation), g is acceleration due to gravity (9.8 m/s2), and h is the height, we can solve for h: 900 J = (1 kg)(9.8 m/s2)h. Solving for h gives us h = 900 J / (9.8 m/s2) = 91.8 m, which we can approximate to 90 m, so the correct answer is (a) 90 m.