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

1 Answer

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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