Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

An object is thrown along a direction inclined at an angle of $45^\circ$ with the horizontal direction. The horizontal range of the particle is equal to:

Vertical height

Twice the vertical height

Thrice the vertical height

Four times the vertical height

Step-by-Step Solution

Formulas: Maximum height ( $H$ ) is given by $H = \frac{v_0^2 \sin^2 \theta}{2g}$ . Horizontal range ( $R$ ) is given by $R = \frac{v_0^2 \sin 2\theta}{g}$ .
Substitute Values: Given $\theta = 45^\circ$ . $H = \frac{v_0^2 (\sin 45^\circ)^2}{2g} = \frac{v_0^2 (1/\sqrt{2})^2}{2g} = \frac{v_0^2 (1/2)}{2g} = \frac{v_0^2}{4g}$ . $R = \frac{v_0^2 \sin(2 \times 45^\circ)}{g} = \frac{v_0^2 \sin 90^\circ}{g} = \frac{v_0^2 (1)}{g} = \frac{v_0^2}{g}$ .
Compare: Comparing $R$ and $H$ : $R = \frac{v_0^2}{g}$ and $H = \frac{1}{4} \frac{v_0^2}{g}$ . Therefore, $R = 4H$ . The range is four times the maximum vertical height.

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