Question

The range of a projectile when fired at \[75{}^\circ \] with the horizontal is \[0.5km\] . What will be its range when fired at \[45{}^\circ \] with the same speed?
(A) \[0.5km\]
(B) \[1.0km\]
(C) \[1.5km\]
(D) \[2.0km\]

Answer 1

Hint: In the given question, we have been asked a question regarding projectile motion. We have been provided with the range of the projectile and the angle from the horizontal at which it was launched. We can break the projectile motion into one-dimensional accelerated motion along the y-axis and one-dimensional motion with uniform speed along the x-axis. Let’s see a detailed solution to the given question.
Formula Used: \[T=\dfrac{2u\sin \theta }{g}\] , \[{{u}_{x}}=u\cos \theta \] , \[R=\dfrac{{{u}^{2}}\sin 2\theta }{g}\]

Step by Step Solution
For the beginning of our solution, since we have been given no particular information, let us assume the initial velocity of the projectile to be \[u\] and the angle at which it is launched be \[\theta \]
Hence, the velocity component of the projectile along the x-axis will be \[u\cos \theta \]
Now, we know that the total time taken by the projectile motion is given as \[T=\dfrac{2u\sin \theta }{g}\]
The range of the projectile can hence be calculated as the product of the horizontal component of the velocity of the projectile and the total time taken for the projectile motion. Mathematically, we can express this as
\[\begin{align}
  & R=u\cos \theta \times \dfrac{2u\sin \theta }{g} \\
 & \Rightarrow R=\dfrac{{{u}^{2}}\sin 2\theta }{g}\left[ \because \sin 2\theta =2\sin \theta \cos \theta \right] \\
\end{align}\]
Let’s consider the first case when the angle with the horizontal is \[75{}^\circ \]
The range will be given as follows
\[\begin{align}
  & {{R}_{1}}=\dfrac{{{u}^{2}}\sin (2\times 75){}^\circ }{g} \\
 & \Rightarrow {{R}_{1}}=\dfrac{{{u}^{2}}\sin (150){}^\circ }{g}--equation(1) \\
\end{align}\]
Similarly, when the projectile is launched at \[45{}^\circ \] with the horizontal, the range can be given as
\[\begin{align}
  & {{R}_{2}}=\dfrac{{{u}^{2}}\sin (2\times 45){}^\circ }{g} \\
 & \Rightarrow {{R}_{2}}=\dfrac{{{u}^{2}}\sin (90){}^\circ }{g}--equation(2) \\
\end{align}\]
Dividing the two equations obtained above, we get
\[\begin{align}
  & \dfrac{{{R}_{2}}}{{{R}_{1}}}=\dfrac{\dfrac{{{u}^{2}}\sin (90){}^\circ }{g}}{\dfrac{{{u}^{2}}\sin (150){}^\circ }{g}}=\dfrac{\sin (90){}^\circ }{\sin (150){}^\circ } \\
 & \Rightarrow \dfrac{{{R}_{2}}}{{{R}_{1}}}=\dfrac{1}{\sin 30{}^\circ }\left[ \because \sin 90{}^\circ =1;\sin (180-\theta )=\sin \theta \right] \\
\end{align}\]
Now, we all know the value of \[\sin 30{}^\circ \] is \[0.5\] and the range \[{{R}_{1}}\] has also been provided to us in the question.
Substituting the values, we get
\[\begin{align}
  & \dfrac{{{R}_{2}}}{0.5km}=\dfrac{1}{0.5} \\
 & \Rightarrow {{R}_{2}}=1.0km \\
\end{align}\]

Hence we can say that option (B) is the correct answer.

Note
We could directly use the final expression obtained for the range of a projectile. We have shown the complete procedure for a better understanding. Also, you should be aware that for projection at \[45{}^\circ \] , the range of a projectile is maximum. The time taken for the projectile motion has been calculated by applying newton’s first law of motion along the y-axis.