Question

A body is projected at a speed of $30\,m\,{s^{ - 1}}$from a very high tower. What will be the speed after $4\,s$ ?
A. $20\,m\,{s^{ - 1}}$
B. $50\,m\,{s^{ - 1}}$
C. $54\,m\,{s^{ - 1}}$
D. $70\,m\,{s^{ - 1}}$

Answer 1

This is an example of horizontal projectile motion. At every point throughout the motion the velocity vector can be represented with its x and y components. Since the acceleration is acting only downwards, the x component of the velocity should not change. In the y direction, we can simply apply the speed equation to get the velocity along the y axis.
The speed equations are
$v = u + at$
$s = ut + \dfrac{1}{2}a{t^2}$
$2as = {v^2} - {u^2}$
Where u is the initial velocity, v is the final velocity, s is the distance covered, t is the time taken and a is the acceleration.
Also, the final velocity is the resultant of the two components. Since the components here are along the x axis and the y axis, the angle between them is ${90^0}$ . And so, the resultant can be calculated as
$v = \sqrt {v_x^2 + v_y^2}$

Complete step by step solution:
The situation can be visualised as

Let the velocity after 4 s be v and its x and y components be ${v_x}$ and ${v_y}$ respectively.
Since there is no acceleration acting in the horizontal direction the velocity in x direction remains same as the initial velocity.
So, ${v_x} = 30\,m\,{s^{ - 1}}$
Now in y direction,
$u = 0$ , $t = 4s$ ,$a = + g$ and ${v_y} = ?$
Using the speed equation $v = u + at$
Substituting the values, we get
${v_y} = 0 + 4g$
$\Rightarrow {v_y} = 4 \times 10$
$\Rightarrow {v_y} = 40\,m\,{s^{ - 1}}$
Now we have found the components of the velocity in both x and y directions.
The resultant of these components is the actual answer.
Since the angle between both the components is ${90^0}$ ,
The resultant is given by$v = \sqrt {v_x^2 + v_y^2}$
Substituting the values,
$v = \sqrt {{{30}^2} + {{40}^2}}$
$\Rightarrow v = \sqrt {1600 + 900}$
Further solving this,
$\Rightarrow v = \sqrt {2500}$
And hence $v = 50\,m\,{s^{ - 1}}$
Hence option B is correct.

Note: Always keep a note of the direction of the acceleration. If it is the same as the motion of the object then we take + sign for acceleration. Likewise, if the direction is opposite then we take a – sign. This is also known as retardation. We use the value of g as $10\,m\,{s^{ - 2}}$ to simplify our calculations.