Question

The variation of horizontal and vertical distances with time is given by $y = 8t - 4.9{t^2}$ m and $x = 6t$m, where t is in seconds. Then the velocity of projection is
A) 8 m/s
B) 6 m/s
C) 10 m/s
D) 14 m/s

Answer 1

Hint:Here we have to calculate the velocity, which is a vector quantity. The motion here is classified in two directions x and y. This motion then has to be added together to form the velocity of projection of the body. Thus, we can calculate the result by using differentiation.

Formula Used:
$v = \sqrt {{v_x}^2 + {v_y}^2} $, where v= velocity of the body. ${v_x}$= velocity of the body in the x direction and ${v_y}$= velocity of the body in the y direction.

Complete step by step answer:
Given in the question,
 $x = 6t$.
Now, differentiating this equation with respect to time t we get,
$\dfrac{{dx}}{{dt}} = 6 = {v_x}$= x – component of the initial velocity.
Again, $y = 8t - 4.9{t^2}$.
Differentiating this equation with respect to time t we get,
$\dfrac{{dy}}{{dt}} = 8 - 9.8t$.
At time t = 0,
$\dfrac{{dy}}{{dt}} = 8 = {v_y}$= y - component of initial velocity.
Now, total velocity,
 $\left| V \right| = \sqrt {{v_x}^2 + {v_y}^2} $.
Putting the values of ${v_x}$ and ${v_y}$ in this equation we get,
$\left| V \right| = \sqrt {{6^2} + {8^2}} $.
Further equating we get,
$\left| V \right| = \sqrt {{6^2} + {8^2}} = \sqrt {36 + 64} = \sqrt {100} = 10$.
Thus, the answer is 10m/s.

So, option (C) is the right answer.

Note:Alternate method of solution. In this method, we are looking at a different approach to solving the same problem. In this method we are not using differentiation but rather diving the velocity into its components in x and y direction.
Let us consider that the body is projected at an angle $\theta $. If we divide the velocity into its components $v\cos \theta $ and $v\sin \theta $.
Then we can rewrite the given equations as
$x = v\cos \theta t = 6t$ and $y = v\sin \theta t - 21g{t^2} = 8t - 4.9{t^2}$
Comparing these equations, we can get,
$v\cos \theta = 6$ and
$v\sin \theta = 8$
Now, $v = \sqrt {v\cos {\theta ^2} + v\sin {\theta ^2}} $
Putting the values of $v\cos \theta $and $v\sin \theta $ in this equation we get,
$v = \sqrt {{6^2} + {8^2}} = \sqrt {36 + 64} = \sqrt {100} = 10$.
Thus, the answer is 10m/s. So, option (C) is the right answer.