A boy playing on the roof of a $10$ m high building throws a ball with a speed of $10$ $m s^{- 1}$ at an angle of $30^{o}$ with the horizontal. How far from the throwing point with the ball be at the height of $10$ m from the ground? $(g = 10 m s^{- 2}, sin 30^{o} = 1 / 2, cos 30^{o} =$ .

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Solution

$A n s w e r i s [5 (3)^{0} .5]$
As when the ball is at a height of 10m from ground it is at same level as of the point of projection so the time taken will be 1 seconds and the distance will be 10cos30*1

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A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 ms−1 at an angle of 30o with the horizontal. How far from the throwing point with the ball be at the height of 10 m from the ground? (g=10ms−2,sin30o=1/2,cos30o=.

Answeris[5(3)0.5]As when the ball is at a height of 10m from ground it is at same level as of the point of projection so the time taken will be 1 seconds and the distance will be 10cos30*1

A boy playing on the roof of a $10$ m high building throws a ball with a speed of $10$ $m s^{- 1}$ at an angle of $30^{o}$ with the horizontal. How far from the throwing point with the ball be at the height of $10$ m from the ground? $(g = 10 m s^{- 2}, sin 30^{o} = 1 / 2, cos 30^{o} =$ .

$A n s w e r i s [5 (3)^{0} .5]$
As when the ball is at a height of 10m from ground it is at same level as of the point of projection so the time taken will be 1 seconds and the distance will be 10cos30*1