Question

a body of mass 1 kg is dropped from top of tower of 50m height what will be KE after 10 m

Answer 1

Answer:

v² = u² + 2as

Therefore, velocity after 10 m will be

» v² = 2 (9.8) 10

» v = √196 = 14 m/s

• K.E. = ½ mv²

Therefore, kinetic energy after falling 10 m will be,

» K.E. = ½ (1) (14)²

» K.E. = 98 J