Question

Two stones are thrown towards each other simultaneously, one vertically downwards from the top of the tower of height 360 m with velocity 10 m/s and other from the ground vertically upwards with velocity 50 m/s. Find the time and the distance from the ground at which they collide. [Take, g = 10 m/s2]
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Answer 1

Answer:

While both the stones are in flight, a

1

=g and a

2

=g

So a

rel

=0⇒V

rel

=constant

⇒X

rel

=(const)t

⇒Curve ofx

rel

v/s t will be straight line.

After the first particle drops on ground, the seperation (x

rel

) will decrease parabolically (due to gravitational acceleration),

and finally becomes zero.

and V

rel

=slope of x

rel

v/st

Answer 2

Given :-

▪ Two stones are thrown towards each other simultaneously. One vertically downwards from the top of the tower of height h = 360 m with velocity v₁ = 10 m/s and other from the ground vertically upwards with velocity v₂ = 50 m/s.

▪ Acceleration due to gravity, g = 10 m/s²

To Find :-

▪ Time taken and the distance from the ground at which they will collide.

Solution :-

Here, We take every measurements in upward direction positive while measurements in downward direction negative.

From these, we have

• g = -10 m/s²
• v₁ = - 10 m/s²

Now, Let us find the relative velocity, since they are in opposite direction, hence the velocities would be added. (Including gravity)

⇒ v = v₁ + v₂

⇒ v = 10 + 50

⇒ v = 60 m/s

The time taken to cover the distance 360 m with this velocity is actually the time at which it will collide.

⇒ Time taken, t = 360 / 60

⇒ t = 36/6

⇒ t = 6 s

So, Let us find the distance from the ground at which they will collide.

⇒ s = ut + 1/2 gt²

⇒ s = 50 × 6 + 1/2 × -10 × 6²

⇒ s = 300 - 5×36

⇒ s = 300 - 180

⇒ s = 120 m

Hence, They will collide after 6 s and at a distance of 120 m from the ground.