Question

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.​

Answer 1

Answer :-

→ The two stones meet after 4 seconds .

→ They meet at a distance of 80 metres . [from the tower].

Explanation :-

Let the two stones meet after a time 't' .

For the 1st stone :-

Let distance travelled by the stone be 'x' metre from the tower .

h = x, u = 0 m/s, g = 10 m/s²

By using the 2nd equation of motion :-

⇒ h = ut + ½gt²

⇒ x = 0(t) + ½ × (10) × t²

⇒ x = 5t² ---(1)

For the 2nd stone :-

Distance travelled by it will be (100 - x) m from the ground .

h = (100 - x) , u = 25 m/s , g = -10 m/s²

Again by the 2nd equation of motion :-

⇒ h = ut + ½gt²

⇒ (100 - x) = 25(t) + ½(-10)t²

⇒ 100 - x = 25t - 5t² ----(2)

______________________________

Putting value of 'x' in eq.2, we get :-

⇒ 100 - 5t² = 25t - 5t²

⇒ 100 = 25t - 5t² + 5t²

⇒ 100 = 25t

⇒ t = 100/25

⇒ t = 4 sec

Now, putting value of 't' in eq.1 :-

⇒ x = 5(4)²

⇒ x = 5(16)

⇒ x = 80 m

Answer 2

Given :-

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s.

To Find :-

When and where the two stones will meet.​

Solution :-

Let the time taken be t

Stone 1

s = ut + 1/2 × at²

-s = (0)(t) + 1/2 × (-10) × t²

-s = 0 + 1/2 × -10t²

-s = 0 + -10t²/2

s = 5t²

Stone 2

100 - s = (25)(t) + 1/2 × (-10)(t²)

100 - s = 25t + 1/2 × -10t²

100 - s = 25t - 5t²

Add both

s + 100 - s = 5t² + 25t - 5t²

100 = 25t

100/25 = t

4 = s

Put the value of s in 1

5t²

5(4)²

5(16)

80 m