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  • The range of a projectile when fired at.  to the horizontal is <img alt="0.5 \mathrm{~km}" src=

The range of a projectile when fired at. 75^{\circ} to the horizontal is 0.5 \mathrm{~km}. What will be its range when fired at 45^{\circ} with the same speed?

Option: 1

0.5 \mathrm{~km}


Option: 2

1.5 \mathrm{~km}


Option: 3

1.0 \mathrm{~km}


Option: 4

2.0 \mathrm{~km}


Answers (1)

best_answer

Problem based on projectile motion.
Range \mathrm{(R)=\frac{u^2 \sin 2 \theta}{g}}
Case 1,\mathrm {R=0.5 \mathrm{~km}, \quad \theta=75^{\circ}}
\mathrm{0.5=\frac{u^2 \sin 150^{\circ}}{g}-\text { (1) }}
\: case 2,\mathrm{\quad R= ?, \theta=45^{\circ}} \mathrm{R=\frac{u^2 \sin 90^{\circ}}{g}}
 To calculate range Equation (1) divide by (2) 
\mathrm{\begin{aligned} & \frac{0.5}{R}=\frac{u^2 \sin 150}{g} \times \frac{g}{u^2 \sin 90^{\circ}} \\ & \frac{0.5}{R}=\frac{\sin 150^{\circ}}{\sin 90^{\circ}} \end{aligned}}
\mathrm{\begin{aligned} & \frac{0.5}{R}=\frac{1 / 2}{1} \\ & 0.5=R / 2 \Rightarrow R=0.5 \times 2 \end{aligned}}
\mathrm{R=1.0 \mathrm{~km}}
 

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shivangi.bhatnagar

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