Question 17 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.
Class 9 - Science - Gravitation Page 144
Answers
"S" = distance travelled by the stone dropped from the top of tower
(100-S) = distance travelled by the projected stone.
◆ i) For stone dropped from the top of tower
-S = 0 + 1/2 (-10) t²
or, S = 5t²
◆ ii) For stone projected upward
(100 - S) = 25t + 1/2 (-10) t²
= 25t - 5t²
Adding i) and ii) , We get
100 = 25t
or t = 4 s
Therefore, Two stones will meet after 4 s.
◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get
S = 5 × 16
= 80 m.
Thus , both the stone will meet at a distance of 80 m from the top of tower.
ANSWER :
Height ( h ) = 100m
Time ( t ) = ?
g = 10 m/s²
Height covered by the falling stone = s₁
Therefore, s₁ = ut + 1/2gt²
s₁ = 0 × t + 1/2 ( 10 ) t²
s₁ = 5t² ---------------------------------- ( 1 )
Let ,The distance covered by the stone thrown upward = s₂
g = ₋ 10 m/s
u = 25 m/s
s₂ = ut + 1/2 gt²
Therefore, s₂ = 25t + 1/2 ( - 10 ) t²
s₂ = 25t -5t² ------------------------------------- ( 2 )
Total height given = 100m
∴ s₁ + s₂ = 100m
5t² + ( 25t - 5t² ) = 100m
∴ 25t = 100m
t = 100/25 = 4 seconds
----------------------------- ( 3 )
Putting the value of equation ( 3 ) in equation ( 1 ), we get
∴ s₁ = 5 t²
= 5 × ( 4 )²
= 80m
Therefore, The two stones will meet after 4 seconds when the falling stone has covered a height of 80m.