A stone is dropped from a height of 50m on earth. At the same time, another stone is thrown vertically upwards from the ground with a velocity up wards from the ground with a velocity of 50m/s. At what height from the ground will the two stones meet (g = -10 m/s2)
Answers
Let the stone which is dropped downwards be stone 1 and the stone which is thrown vertically upwards be stone 2.
Given:-
- Initial velocity of stone 1 = u₁ = 0 m/s
- Initial velocity of stone 2 = u₂ = 50 m/s
- Height = 50 m
- g = 10 m/s²
**NOTE: Acceleration due to gravity = g will be always equal to +10 m/s². If you want to show -10 m/s² (when it's acting downwards), then you can use (-g), but g will always remain +10 m/s².
To find:-
- Distance above the ground where the two stone will meet.
Answer:-
▪Let the distance above the ground at which the stones will meet be x.
▪Then the distance between the point where stone 1 is released and the point where the stones meet will be (50 - x).
▪Time taken for both the balls to reach the point will be same. So t₁ = t₂ = t.
Refer to the attachment.
▪For Ball 1:-
s₁ = u₁t + (1/2)a₁t²
Sign convention: s₁ = -(50 - x), since displacement is in downward direction. a₁ = (-g) since it's acting in downward direction. Putting available values,
→ -(50 - x) = (0 × t) + [1/2 × (-10) × t²]
→ -(50 - x) = -[1/2 × 10 × t²]
→ 50 - x = 5t² ----( 1 )
▪For Ball 2:-
s₂ = u₂t + (1/2)a₂t²
Sign convention: s₂ = x, since displacement is in upward direction. a₂ = (-g) since it's acting in downward direction. Putting available values,
→ x = (50 × t) + [1/2 × (-10) × t²]
→ x = 50t - 5t² ----( 2 )
Putting the value of 5t² from ( 1 ) in ( 2 ),
x = 50t - (50 - x)
→ x = 50t - 50 + x
→ 50 = 50t
→ t = 1 second ----( 3 )
Putting the value of ( 3 ) in ( 1 ),
50 - x = 5(1)²
→ 50 - x = 5
→ x = 45 metres.
Distance between ground and the point where the two stones meet = x
So,
Distance between ground and the point where the two stones meet = 45 metres. Ans.
Answer:
plz refer to the attachment
Explanation:
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