A ball is projected from the ground with a speed 15 ms$-$1 at an angle $\theta$ with horizontal so that its range and maximum height are equal, then 'tan $\theta$' will be equal to :

Question

A ball is projected from the ground with a speed 15 ms^$-$1 at an angle $\theta$ with horizontal so that its range and maximum height are equal,
then 'tan $\theta$' will be equal to :

${1 \over 4}$

${1 \over 2}$

Solution

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