Question

prove that time of ascent = time of decent

Answer 1

During acsent: g=9.8 , t1=time taken for ascent , u=initial velocity , v=final velocity=0 Therefore, t1= (v-u)/-g= (0-u)/g = u/g During descent, g=9.8 , u=initial velocity=0 , v=final velocity, t2=time taken for the descent Therefore, t2= (v-u)/g= (v-o)/g= v/g The velocity with which the body is thrown = The velocity with which the body hits the ground. (experimented truth) Therefore, u/g=v/g Thus, t1=t2 or, the time of ascent = the time of descent (hence proved)