A ball is shot in a long hall having a roof at a height of $15$ m with speed of $25$ m/s at an angle of $53^{o}$ with the floor. The ball lands on the floor at a distance shown x $=$ ________ m from the point of projection.(Assume collisions as elastic if any)

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Solution

The distance covered by the ball is nothing but the horizontal range of the ball which is mathematically expressed as R $= \frac{v_{0}^{2} sin 2 θ}{g}$ ------(A)
where $v_{0}^{2}$ =speed of the ball=25 m/s;g=acceleration due to gravity=9.81 m/ $s^{2}$ and $θ$ =angle of projectile= $53^{0}$
Now putting the above mentioned values in A we get
R $= \frac{{(25)}^{2} sin (2 \times 53)}{9.81} = \frac{625 \times sin 106}{9.81} = 61.24 m$

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A ball is shot in a long hall having a roof at a height of 15 m with speed of 25 m/s at an angle of 53o with the floor. The ball lands on the floor at a distance shown x=________ m from the point of projection.(Assume collisions as elastic if any)

A ball is shot in a long hall having a roof at a height of $15$ m with speed of $25$ m/s at an angle of $53^{o}$ with the floor. The ball lands on the floor at a distance shown x $=$ ________ m from the point of projection.(Assume collisions as elastic if any)