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Byju's Answer

Standard IX

Physics

Derivation of Position-Time Relation by Graphical Method

Two balls A...

Question

Two balls $A$ and $B$ are simultaneously thrown. $A$ is thrown from the ground level with a velocity of $20 m s^{- 1}$ in the upward direction and $B$ is thrown from a height of $40 m$ in the downward direction with the same velocity. Where will the two balls meet?

15 m

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25 m

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35 m

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45 m

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Solution

The correct option is B $15 m$
Let both the ball meet after time $t$ and $x$ be the distance covered by ball $A$ .
So $, x = 20 t - \frac{1}{2} (10) t^{2} . . . (1)$
For ball $B, (40 - x) = 20 t + \frac{1}{2} (10) t^{2} . . . (2)$
Adding $(1) & (2), 40 = 40 t o r t = 1 s$
so $x = 20 - 5 = 15 m$

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Two balls A and B are simultaneously thrown. A is thrown from the ground level with a velocity of 20 ms−1 in the upward direction and B is thrown from a height of 40 m in the downward direction with the same velocity. Where will the two balls meet?

The correct option is B 15 mLet both the ball meet after timet and x be the distance covered by ball A.So, x=20t−12(10)t2...(1)For ball B,(40−x)=20t+12(10)t2...(2) Adding (1) & (2),40=40tort=1 s so x=20−5=15 m

Two balls $A$ and $B$ are simultaneously thrown. $A$ is thrown from the ground level with a velocity of $20 m s^{- 1}$ in the upward direction and $B$ is thrown from a height of $40 m$ in the downward direction with the same velocity. Where will the two balls meet?

The correct option is B $15 m$
Let both the ball meet after time $t$ and $x$ be the distance covered by ball $A$ .
So $, x = 20 t - \frac{1}{2} (10) t^{2} . . . (1)$
For ball $B, (40 - x) = 20 t + \frac{1}{2} (10) t^{2} . . . (2)$
Adding $(1) & (2), 40 = 40 t o r t = 1 s$
so $x = 20 - 5 = 15 m$