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?

Asked by Abhisek | 1 year ago |  94

##### Solution :-

Maximum horizontal distance, R = 100 m

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45° i.e.,

$$\theta$$ = 33.60°

The horizontal range for a projection velocity v, is given as:

R =$$\dfrac{u^{2}\sin 2\theta}{g}100$$

= $$\cfrac{u^{2}}{g}\sin 90^{\circ}\dfrac{u^{2}}{g} = 100$$—— (i)

The ball will achieve the max height when it is thrown vertically upward. For such motion, the final velocity v is 0 at the max height H.

Acceleration, a = – g

Using the 3rd equation of motion:

$$v^{2} – u^{2} = -2gH = \dfrac{1}{2} \times \frac{u^{2}}{g}$$

=$$H = \dfrac{1}{2} \times 100$$ = 50 m

Answered by Pragya Singh | 1 year ago

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