Question:
the distance travelled by a freely 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:
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.
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