58) Two balls, A and B are simultaneously projected from the top of a building at 10 mis upwards & 20 m/s downwards respectively. Find
out the separation btwn them 3 sec after projection.

90 m

Explanation:

For ball A::

The equation of a ball moving upwards is given by the formula:

s = ut - (1/2)gt²

where u is the initial velocity, t is the time covered, g is the acceleration due to gravity = 10 m/s² and s is the distance travelled.

Given that:

u = 10 m/s, t = 3 sec, g = 10 m/s², hence:

s = (10 * 3) - (1/2) * 10 * 3²

s = 30 - 45

s = -15 m

The negative sign means the object travels in the opposite direction. Hence s = 15 m

For ball B::

The equation of a ball falling downwards is given by the formula:

s = ut + (1/2)gt²

Given that:

u = 10 m/s, t = 3 sec, g = 10 m/s², hence:

s = (10 * 3) + (1/2) * 10 * 3²

s = 30 + 45

s = 75 m

Hence the separation between the two balls after 3 sec of projection = 15 m + 75 m = 90 m

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

prolly a beacuse the momentuem the car has increases while going down hill