4.16 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

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Vishal Baghel | Contributor-Level 10

8 months ago

4.16 We know for a projectile motion, horizontal range is given by

R = $\frac{u^{2}}{g} s i n 2 θ$ , where R is the horizontal range, u is the velocity and $θ$ is the angle of projectile

So $\frac{u^{2}}{g}$ = 100/sin 2 $θ$

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection $θ$ = 45 $°$ , $\frac{u^{2}}{g}$ = 100

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