Question

Find the initial velocity of the projectile, ${V_0}$, if a projectile is fired into the air from the edge of a $100m$ high cliff at an angle of ${37^ \circ }$ above the horizontal and the projectile hits a target $400m$ away from the base of the cliff. (Neglect the air friction and assume x – axis to be horizontal and y – axis to be vertical).
A) ${V_0} = 25\sqrt 5 m/\sec $
B) ${V_0} = 23\sqrt 3 m/\sec $
C) ${V_0} = 22\sqrt 2 m/\sec $
D) ${V_0} = 27\sqrt 7 m/\sec $

Answer 1

-Hint:-Take the distance and acceleration of projectile in x – axis as positive and in y – axis as negative. Find the values of acceleration of x – axis and y – axis respectively. Now use the values of acceleration, distance and velocity in this formula of projectile –
$S = ut + \dfrac{1}{2}a{t^2}$
where, $S$ is the distance travelled by projectile,
$t$ is the fight time,
$a$ is the acceleration of a projectile.

Complete Step by Step Solution:-
Let ${S_x},{U_x}$ and ${a_x}$ be the distance, velocity and acceleration of the horizontal axis or x – axis of the projectile
So, according to the question, it is given that
$
  {S_x} = 400m \\
  {U_x} = {V_0}\cos {37^ \circ } \Rightarrow \dfrac{4}{5}{V_0} \\
 $
After hitting the target in the horizontal axis, the acceleration of the projectile becomes zero.
$\therefore {a_x} = 0$
Now, using the formula –
${S_x} = {U_x}t + \dfrac{1}{2}a{t^2}$
Putting the values in the above formula
$400 = \dfrac{4}{5}{V_0}t + \dfrac{1}{2} \times 0{t^2}$
By solving, we get
$t = \dfrac{{500}}{{{V_0}}} \cdots (1)$

Now, let ${S_y},{U_y}$ and ${a_y}$ be the distance, velocity and acceleration of the vertical axis or y – axis of the projectile
For the vertical axis, the distance and acceleration will be taken negative.
According to the question, we get –
$
  {S_y} = - 100m \\
  {U_y} = {V_0}\sin {37^ \circ } \Rightarrow \dfrac{3}{5}{V_0} \\
 $
Due to the vertical axis, the acceleration is equal to the acceleration due to gravity
$\therefore {a_y} = - 10m/{\sec ^2}$
Now, using these values in the formula –
 ${S_y} = {U_y}t + \dfrac{1}{2}a{t^2}$
$
   - 100 = \dfrac{3}{5}{V_0}t - \dfrac{1}{2} \times 10 \times {t^2} \\
   - 100 = \dfrac{3}{5}{V_0}t - 5{t^2} \cdots (2) \\
 $
Using the value of $t$ from equation $(1)$ in equation $(2)$, we get the value of flight time as –
$t = \dfrac{{20}}{{\sqrt 5 }}\sec $
Now, putting this value of flight time in equation $(1)$, we get
$
  \dfrac{{20}}{{\sqrt 5 }} = \dfrac{{500}}{{{V_0}}} \\
  {V_0} = 25\sqrt 5 m/\sec \\
 $
Therefore, the correct option is (A).

Note:-When an object is thrown or is at flight then it is said to be projectile and the motion of object is called Projectile motion. The path of the projectile is parabolic. The projectile motion depends upon the flight time, maximum height and horizontal range of projectile.