A projectile is launched from the top of a 40 m high cliff with an initial speed of 50 m/s at an unspecified angle. Determine its speed upon striking the ground.

Step-by-step solution

Using the principle of conservation of mechanical energy, the total energy of the projectile remains constant. Let m be the mass of the projectile. Taking the ground as the reference level for potential energy: Initial energy E_i = 1 2 mu^2 + mgh Final energy E_f = 1 2 mv^2 Equating initial and final energies: 1 2 mv^2 = 1 2 mu^2 + mgh v^2 = u^2 + 2gh Substituting the given values ( u = 50 m/s , h = 40 m ) and taking g = 10 m/s ^2 : v^2 = (50)^2 + 2(10)(40) v^2 = 2500 + 800 = 3300 v = 3300 57.4 m/s The closest valu