If a ball is dropped from a high building and it reaches the ground in 5 seconds. Find the velocity with which it hits the ground? What is its average speed? What is the height of the building? 

Ans. (60 m s⁻¹,  30 m s⁻¹, 180 m, )


Given data:


Initial Velocity of the Cricket Ball = →vi= 0 m s⁻¹

Time of the ball during which it falls to the ground  = t = 6 s

Gravitational acceleration = g = 10 m s⁻² (+ve for downwards motion)


To Find:

Final Velocity of the Cricket Ball = →vf= ? m s⁻¹

Average Speed = vave = ?

Height of the building = h = ? m


Solution:  



Here, we are using the 1st equation of motion

→vf = →vi + gt

by putting values

→vf   = 0 ms⁻¹ + 10 ms⁻²ｘ 6 s

by simplifying we get

→vf   = 60 ms⁻¹ 

Thus, the velocity with which the ball hit the ground  is 60 ms⁻¹




To find average speed by equation vave = ht. But we need to find first the height of the building h

To calculate the height of the building we are using the 2nd equation of motion for veridical motion where the replacement of (a with g) and (S with h) so,

h = vi t  + 12gt²

by putting vales

h = o ms⁻¹ｘ 6 s + 12(10 m s⁻²) (6 s)²

h = 180 m

So, the height of the building is 180 meters. 




Now we know that 

vave = ht

by putting vales

vave = 180m6s

vave = 30 m/s ------------------Ans.

Or by using the equation (2nd method)

vave = vi+vf2

by putting vales

vave = o+602 m/s = 30 m/s ------Ans. 




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