Answer:
Change in mechanical energy, ![\Delta E=283.2\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D283.2%5C%20J)
Explanation:
It is given that,
Mass of the projectile, m = 12 kg
Speed of the projectile, v = 20 m/s
Maximum height, h = 18 m
Initially, the projectile have only kinetic energy. it is given by :
![K=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![K=\dfrac{1}{2}\times 12\ kg\times (20\ m/s)^2](https://tex.z-dn.net/?f=K%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%2012%5C%20kg%5Ctimes%20%2820%5C%20m%2Fs%29%5E2)
K = 2400 J
Finally, it have only potential energy. it is given by :
P = mgh
![P=12\ kg\times 9.8\ m/s^2\times 18\ m](https://tex.z-dn.net/?f=P%3D12%5C%20kg%5Ctimes%209.8%5C%20m%2Fs%5E2%5Ctimes%2018%5C%20m)
P =2116.8 J
The change in mechanical energy is given by :
![\Delta E=K-P](https://tex.z-dn.net/?f=%5CDelta%20E%3DK-P)
![\Delta E=2400-2116.8](https://tex.z-dn.net/?f=%5CDelta%20E%3D2400-2116.8)
![\Delta E=283.2\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D283.2%5C%20J)
So, the change in mechanical energy is 283.2 J. Hence, this is the required solution.