Question

the body thrown vertically down from a height moves 40 m during 4 th second of its motion. determine it initial speed​

Answer 1

Answer:

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.

Acceleration due to gravity, g = 10m/s

2

(downward motion).

Maximum height, s = H.

As the body is thrown upward a = -g the relation v

2

=u

2

−2as gives v

2

=u

2

−2aH, we have,

H =

2g

u

2

−v

2

=

2(10m/s

2

)

40m/s

2

−0

2

=

20

1600

=80m.

If a stone is thrown vertically upward, it returns to its initial position after achieving maximum height.

so, the net displacement = Difference of positions between initial and final positions = 0.

Total distance covered = 80 m + 80 m = 160 m.

Hence, the displacement is 0 and the total distance covered is 160 m.