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.

Draw diagram please, 9th Class... ​

Answer 1

Solution :

Here, Height of the tower is 100 m. Now, suppose the two stones meet at a point P which is at height x above the ground, so that the distance of point P from the top of the tower is 100 - x (Refer to the attachment)

i) For the stone falling from top of tower :

• Height (h) = (100 - x) m

• Initial Velocity (u) = 0

• Time (t) = ?

• Acceleration due to gravity (g) = 9.8 m/s² = (10 m/s²) Approx

Now,

★ h = ut + ½ gt²

→ x = 0(t) + ½ × 10 × (t)²

→ x = 5 t².... i)

ii) For stone projected vertically upwards :

• Height (h) = x m

• Initial Velocity (u) = 25 m/s

• Time (t) = ?

• Acceleration due to gravity (g) = - 9.8 m/s² = ( - 10 m/s) Approx

Now,

★ s = ut + ½ at²

→ 100 - x = 25 × t + ½ × (- 10) × t²

→ 100 - x = 25t - 5 t²......ii)

Adding i) and ii)

→ 100 - x + x = 5t ² + 25t - 5t²

→ 100 = 25t

→ t = $\sf\cancel\dfrac{100}{25}$

→ t = 4 seconds

$\therefore$ Two stones will meet after a time of 4 seconds.

Now, we've to find the height at which two stones will meet

From i) we've :

★ x = 5t²

Putting the value of t = 4

→ x = 5(4)²

→ x = 5(16)

→ x = 80 m

$\therefore$ Two stones will meet at the height of 80 metres from the ground.