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Show that for angles of projection theta...

Show that for angles of projection `theta` and `90^(@)-theta`, the range of the projectile is the same.

Text Solution

Verified by Experts

We know that `R=(v_0^2sin2theta)/(g)` Put `theta=90^(@)-theta, R_1=(v_0^2)/gsin2(90^(@)-theta)`
i.e. `R_1=(v_0^2)/gsin(180^(@)-2theta)`. `R_1=(u^2)/gsin2theta`
`:." "R=R_1`
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