Question

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.
​

Answer 1

90 m

Step-by-step 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