Question 55: Motion in 2D & Vectors

Question

Two objects are projected with same velocity 'u' however at different angles $$\alpha$$ and $$\beta$$ with the horizontal. If $$\alpha+\beta=90^\circ$$, the ratio of horizontal range of the first object to the 2nd object will be :

Accepted Answer

$$
	ext {Range}=\frac{u^2 \sin 2 	heta}{g}
$$
Range for projection angle " $\alpha$ "
$$
\mathrm{R}_1=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}
$$
Range for projection angle " $\beta$ "
$$
\begin{aligned}
& \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \beta}{\mathrm{g}} \\
& \alpha+\beta=90^{\circ}(	ext { Given }) \\
& \Rightarrow \beta=90^{\circ}-\alpha \\
& \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2\left(90^{\circ}-\alphaight)}{\mathrm{g}} \\
& \mathrm{R}_2=\frac{\mathrm{u}^2 \sin \left(180^{\circ}-2 \alphaight)}{\mathrm{g}} \\
& \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}} \\
& \Rightarrow \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}ight)}{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}ight)}=\frac{1}{1}
\end{aligned}
$$

Answer

1 : 1

Answer

2 : 1

Answer

1 : 2

Answer

4 : 1

Motion in 2D & Vectors

Two objects are projected with same velocity 'u' however at different angles $\alpha and \beta with the horizontal. If \alpha+\beta=90^\circ$, the ratio of horizontal range of the first object to the 2nd object will be :

Choose the correct answer: