An object is thrown along a direction inclined at an angle of $45^\circ$ with the horizontal direction. The ho

Question

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:

Accepted Answer

1.  **Formulas:** 
    *   Maximum height ($H$) is given by $H = \frac{v_0^2 \sin^2 	heta}{2g}$ .
    *   Horizontal range ($R$) is given by $R = \frac{v_0^2 \sin 2	heta}{g}$ .
2.  **Substitute Values:** Given $	heta = 45^\circ$.
    *   $H = \frac{v_0^2 (\sin 45^\circ)^2}{2g} = \frac{v_0^2 (1/\sqrt{2})^2}{2g} = \frac{v_0^2 (1/2)}{2g} = \frac{v_0^2}{4g}$.
    *   $R = \frac{v_0^2 \sin(2 	imes 45^\circ)}{g} = \frac{v_0^2 \sin 90^\circ}{g} = \frac{v_0^2 (1)}{g} = \frac{v_0^2}{g}$.
3.  **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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