Question

33. A stone is dropped from top of a tower 100m height, At the same instant another stone is
thrown vertically upward from the base of the tower with velocity of 25ms". When and
where will the two stones meet.​

Answer 1

Answer:

20m

Explanation:

A stone dropped from the top of a tower of height 100m . Let time taken to meet the another stone be 't' . Displacement covered by stone during this time.

S1 = ut + 1/2 at²

=0×t + 1/2 × 10 × t²

S1 = 5t²

Another stone is thrown from the base of the tower with velocity of 25m/s . Let the displacement be

S2 = ut + 1/2 at²

= 25×t - 1/2 ×10 × t²

S2 = 25t - 5t²

For collision

S1 + S2 = 100

5t² + 25t - 5t² =100

25t = 100

t = 4 sec

Putting t = 4 in S1

S1 = 5t²

= 5 (4)²

= 5 (16)

S1 =80m

Height - collision = time where two stones will meet

100 - 80 = 20 m

They will meet at 20m height from the base of the tower.