Question

If a ball free falling in a velocity of 5.8 m/s and height of 7.4 which ball will fall faster the ball falling at the equator or at the polar region and what is the difference in their time?​

Answer 1

correct answer is 34269

Answer 2

$\sf\huge\bold{Answer:}$

Given :

• Initial velocity of body = 5.8m/s
• Height = 7.4m

To find :

→Which ball will fall faster

• The ball at equator or at poles.

→the time difference between balls.

Solution :

First of all,

Acceleration due to Gravity

g(at poles)=9.8m/s²

g(at equator)=9.7m/s²

Also,

$\boxed{\tt \red{ acceleration = \frac{change \: in \: velocity}{time} }}$

At equator :

• g = 9.7m/s²
• v = 5.8m/s

→g=v/t

→9.7=5.8/t

→t=5.8/9.7

→t=0.5979sec

→t≈0.6sec

time taken at equator = 0.5979s ≈ 0.6s

At poles :

• g = 9.8m/s²
• v = 5.8m/s

→g=v/t

→9.8=5.8/t

→t=5.8/9.8

→t=0.5918sec

→t≈0.6sec

time taken at poles = 0.5918s ≈ 0.6s

• Difference in time = time at equator-time at poles

With accurate values :

Difference in time = 0.5979s-0.5918s = 0.0061s

With approximate values :

Difference in time ≈ 0.6s-0.6s ≈ 0s

• to find : Which ball will fall faster from a height of 7.9m

Formula used :

[final velocity = initial velocity + acceleration × time]

v = u+at

• At equator

→v = 5.8m/s+(9.7m/s²×0.6s)

→v=5.8+

→v=11.62m/s

• At poles

→v=5.8m/s+(9.8m/s²×0.6s)

→v=5.8+

→v=11.68m/s

It can be clearly seen that 11.68m/s > 11.62m/s

i.e. final velocity at poles > final velocity at equator.

Hence, the ball at poles will fall faster than the ball at equator with a velocity of 11.68m/s.