Question

The distances travelled by a falling body in the last second of its motion, to that in the last but one second is 7:5 , The velocity with which body strikes the ground is

Answer 1

Answer:Let u, v, s1 be for second last second

so v, v1, s2 is for last second

v= u+gt

v= u+g*1 = u+g

s1= ut+ 1/2gt^2

s1= u*1 + 1/2g * 1^2

s1= u + g/2

s2= vt + 1/2gt^2

s2= v+ g/2

s2= (u+g)+ g/2 = u+ 3g/2

so,

{u+3g/2} / {u+g/2} = 7/5

5u + 15g/2 = 7u+ 7g/2

2u = 4g

u = 2g

Now final vel at strike after 2 s

v1 = u+ gt

v1= 2g + g*2

v1= 4g

v1 = 4*9.8

= 39.2 m/s

Thus, the body strikes the ground with a velocity of 39.2 m/s

Explanation: