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Standard XII

Physics

Multiplication with Vectors

Two tall buil...

Question

Two tall buildings are 40 m apart. With what is speed must a ball be thrown horizontally from a window 145 m
above the ground in one building, so that it will enter a window 22.5 m from the ground in the other?

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Solution

The correct option is B
40 = $u \sqrt{\frac{2 (145 - 22.5)}{9.8}}$
$\Rightarrow u = 8 m s^{- 1}$

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Two tall buildings are 40 m apart. With what is speed must a ball be thrown horizontally from a window 145 m above the ground in one building, so that it will enter a window 22.5 m from the ground in the other?

The correct option is B 40 =u√2(145−22.5)9.8 ⇒u=8 ms−1

Two tall buildings are 40 m apart. With what is speed must a ball be thrown horizontally from a window 145 m
above the ground in one building, so that it will enter a window 22.5 m from the ground in the other?

The correct option is B
40 = $u \sqrt{\frac{2 (145 - 22.5)}{9.8}}$
$\Rightarrow u = 8 m s^{- 1}$