Question
A cricketer can throw a ball to a maximum horizontal distance of 100m. How much high above the ground can the cricketer throw the same ball?
✓
Answer
Maximum horizontal distance, R = 100m The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is $45^\circ,\text{i.e.,}\theta=45^\circ.$ The horizontal range for a projection velocity v, is given by the relation: $\text{R}=\text{u}^2\sin2\theta/\text{g}$ $100=\text{u}^2\sin90^\circ/\text{g}$ $\frac{\text{u}^2}{\text{g}}=100\ ...(\text{i})$ The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H. Acceleration, a = -g Using the third equation of motion: $\text{v}^2-\text{u}^2=-2\text{gH}$ $\text{H}=\frac{\text{u}^2}{2\text{g}}=\frac{100}{2}=50\text{m}$
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