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 = `\vec v_i`= 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 = `\vec v_f`= ? m s⁻¹

Average Speed = `\v_{ave}` = ?

Height of the building = h = ? m


Solution:  



Here, we are using the 1st equation of motion

`\vec v_f` = `\vec v_i` + gt

by putting values

`\vec v_f`   = 0 ms⁻¹ + 10 ms⁻²ï½˜ 6 s

by simplifying we get

`\vec v_f`   = 60 ms⁻¹ 

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




To find average speed by equation `\v_{ave}` = `\frac {h}{t}`. 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  + `\frac{1}{2}`gt²

by putting vales

h = o ms⁻¹ï½˜ 6 s + `\frac{1}{2}`(10 m s⁻²) (6 s)²

h = 180 m

So, the height of the building is 180 meters. 




Now we know that 

`\v_{ave}` = `\frac {h}{t}`

by putting vales

`\v_{ave}` = `\frac {180 m}{6 s}`

`\v_{ave}` = 30 m/s ------------------Ans.

Or by using the equation (2nd method)

`\v_{ave}` = `\frac {v_i + v_f}{2}`

by putting vales

`\v_{ave}` = `\frac {o + 60}{2}` m/s = 30 m/s ------Ans. 




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