a ball of mass 100g is projected vertically upwards from the ground with a velocity of 49m/s at the same time another identical ball is dropped from a height of 98m to fall freely along the same path as followed by the first ball. After some time the two balls collide and stick together and finally fall together. Find the time of flight of the masses

Question

a ball of mass 100g is projected vertically upwards from the ground with a velocity of 49m/s at the same time another identical ball is dropped from a height of 98m to fall freely along the same path as followed by the first ball. After some time the two balls collide and stick together and finally fall together. Find the time of flight of the masses

Answer 1

To find the time of flight of the masses, we need to calculate the time it takes for each ball to reach its maximum height.

Let's start by finding the time for the vertically projected ball (Ball A) to reach its maximum height. We can use the following kinematic equation:

vf = vi + at

Where:
vf = final velocity (0 m/s at maximum height)
vi = initial velocity (49 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s^2)

Rearranging the equation to solve for time (t), we have:

t = (vf - vi) / a

Substituting the values:

t = (0 - 49) / -9.8
t = 5 seconds

Thus, it takes 5 seconds for Ball A to reach its maximum height.

Now, let's find the time it takes for the ball dropped from a height (Ball B) to reach the same height. Since Ball B is dropped, its initial velocity is 0 m/s.

Using the same equation as before, we have:

t = (vf - vi) / a

Again, substituting the values:

t = (vf - 0) / -9.8
t = vf / -9.8

To find vf, we can use another kinematic equation:

vf^2 = vi^2 + 2ad

Where:
vi = initial velocity (0 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s^2)
d = displacement (vertical distance traveled)

The displacement, in this case, is 98 meters (height from which Ball B is dropped).

vf^2 = 0^2 + 2(-9.8)(98)
vf^2 = -1960
vf = -44.271 m/s (negative because the velocity is directed downward)

Substituting vf into the equation for time:

t = (-44.271) / -9.8
t = 4.51 seconds (approximately)

Now, we have the time it takes for each ball to reach the same height: 5 seconds for Ball A and 4.51 seconds (approximately) for Ball B.

Since the balls collide at the maximum height, the total time of flight will be the time it takes for Ball A to reach its maximum height (5 seconds).