Question

a ball is dropped from a height of 78.4m , find it speed 2 sec after it fallen , it speed when it has fallen and time to fallen

Answer 1

The distance is 45m. (s)

Acceleration is 9.8 m/s^2 (acceleration due to gravity). (a)

We have to find the time (t)

By the 3rd equation of motion,

s = ut + 1/2 at^2 (here u is the initial velocity of the ball)

Since the initial velocity was 0 m/s, the formula becomes s = 1/2 at^2.

(substituting the values in the formula)

45 = 1/2 x 9.8 x t^2

45 = 9.8/2 x t^2

45 = 4.9 x t^2 (9.8/2 = 4.9)

(divide 4.9 on both sides)

45/4.9 = 4.9 x t^2 / 4.9

9.1836 = t^2

3.030 = t

So it takes approximately 3 seconds for the ball to reach the surface of the earth (neglecting air resistance).

Answer 2

The distance is 45m. (s)

Acceleration is 9.8 m/s^2 (acceleration due to gravity). (a)

We have to find the time (t)

By the 3rd equation of motion,

s = ut + 1/2 at^2 (here u is the initial velocity of the ball)

Since the initial velocity was 0 m/s, the formula becomes s = 1/2 at^2.

(substituting the values in the formula)

45 = 1/2 x 9.8 x t^2

45 = 9.8/2 x t^2

45 = 4.9 x t^2 (9.8/2 = 4.9)

(divide 4.9 on both sides)

45/4.9 = 4.9 x t^2 / 4.9

9.1836 = t^2

3.030 = t

So it takes approximately 3 seconds for the ball to reach the surface of the earth (neglecting air resistance).