The speed of a projectile at the highest point becomes 1/√2 times its initial speed. The horizontal range of the projectile will be - VgyanHub

Question:

The speed of a projectile at the highest point becomes 1/√2 times its initial speed. The horizontal range of the projectile will be

Options :

(a) u2/g
(b) u2/2g
(c) u2/3g
(d) u2/4g

Answer :

(a) u2/g

Answer Steps :

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