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Q. 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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We are given the range of projectile motion.

                                             R\ =\ \frac{u^2\ \sin 2\Theta }{g}

Substituting values :

                                            100\ =\ \frac{u^2\ \sin 90^{\circ} }{g}

So,                                         \frac{u^2 }{g}\ =\ 100

Now since deacceleration is also acting on the ball in the downward direction :

                                            v^2\ -\ u^2\ =\ -2gh

Since final velocity is 0, so maximum height is given by :

                                                 H\ =\ \frac{u^2}{2g}

 or                                             H\ =\ 50\ m

Posted by

Devendra Khairwa

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