A Stone is allowed to fall from the top of a tower 100m 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 stone meet.
Answers
Solution:
Let the distance that the stone travelled from top to down be x.
(i) For stone dropped from the top of tower:
Given:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 10 m/s²
By second equation of motion,
Substituting the values, we get:
x = (0 × t) + ½ × 10 × t²
x = 0 + 5t²
x = 5t² - - - - (1)
(ii) For stone projected upwards:
Distance (s) = (100 - x)
Initial velocity (u) = 25 m/s
By second equation of motion,
Substituting the values, we get:
(100 - x) = (25 × t) + ½ × - 10 × t²
100 - x = 25t - 5t²
- x = 25t - 5t² - 100
- (x) = - (-25t² + 5t² + 100)
x = - 25t² + 5t² + 100 - - - - (2)
From equation (1) and (2),
x = x
5t² = -25t + 5t² + 100
5t² + 25t - 5t² = 100
25t = 100
t = 100/25
t = 4 s
Therefore:
Two stones will meet after 4 s.
Now, putting the value of t in eq (1),
x = 5t²
x = 5 (4)²
x = 5 × 16
x = 80 m
Hence:
The two stones will meet after 4 seconds when the falling stone has covered a height of 80 m.
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Answered by: Niki Swar, Goa❤️