Question

Galileo writes that for angles of projection of a projectile at angle $\left( {45 + a} \right)$ and $\left( {45 - a} \right)$, the horizontal ranges described by the projectile are in the ratio of:
A. $2:1$
B. $1:2$
C. $1:1$
D. $2:3$

Answer 1

Hint: Here first we have to find the range for $\left( {45 + a} \right)$ using the range formula and then we have to find the range for $\left( {45 - a} \right)$. At last we can find the ratio of the ranges with the values.

Complete step by step answer:
Projectile is a body tossed into the vertical plane with an initial velocity and then travels in two dimensions without being driven by any piston or gasoline under the motion of gravity alone. Its acceleration is considered the acceleration of projectiles. A Projectile’s course is called its trajectory.
Projectile motion is a two-dimensional motion case. Any two-dimensional motion case can be resolved into two instances of one-dimensional motion, one along the x-axis and the other along the y-axis. The two instances can be analysed as two instances of one-dimensional motion. The results of two instances can be compared to see the net effect using vector algebra. Vertical and horizontal motions are totally independent of each other.
The horizontal distance travelled by the projectile motion of the body is called the projectile range.
The formula for range is given below-
Range $ = \dfrac{{{u^2}\sin (2a)}}{g}$, where $u = $initial velocity, $g = $ acceleration due to gravity
Now we find the range for $\left( {45 + a} \right)$

Range (1)
 $
   = \dfrac{{{u^2}\sin \left( {2 \times \left( {45 + a} \right)} \right)}}{g} \\
   = \dfrac{{{u^2}\cos 2a}}{g} \\
$

Now we find the range for $\left( {45 - a} \right)$
Range (2)
  $
   = \dfrac{{{u^2}(90 - 2a)}}{g} \\
   = \dfrac{{{u^2}\cos 2a}}{g} \\
$

Hence, $\dfrac{{{\text{Range(1)}}}}{{{\text{Range(2)}}}} = \dfrac{1}{1} = 1:1$

Note:
Here we have to keep in mind that though the formula for range is the same. But the angle may vary in projectile. In the first case the angle is $\left( {45 + a} \right)$ and in the second case the angle is $\left( {45 - a} \right)$. Also the horizontal range is maximum when the angle of projection is ${45^ \circ }$.