Question

The vertical height $y$ and horizontal distance $x$ of a projectile on a certain planet are given by $x = \left( {3t} \right)\;{\rm{m}}$, $y = \left( {4t} \right)\;{\rm{m}}$ where $t$is in seconds. Find the speed of projection (in m/s) ?

Answer 1

Hint: The differentiation of position with respect to time is equal to velocity and the both the velocities are perpendicular to each other. Find the magnitude of both the velocities.

Formula used: The velocity of a body is given by $v = \dfrac{{dx}}{{dt}}$ where $x$ is the position or displacement of the body and $t$ is time.

Complete step by step answer:
From the given question, we know that the vertical height of the projectile is $y = \left( {4t} \right)\;{\rm{m}}$ and the horizontal distance of the projectile is $x = \left( {3t} \right)\;{\rm{m}}$.

We know that the differentiation of position with respect to time is equal to velocity.

The vertical velocity of the projectile is,
${v_y} = \dfrac{{dy}}{{dt}}$
Substitute the expression in the above equation. We get,
$
{v_y} = \dfrac{{d\left( {4t} \right)}}{{dt}}\\
 = 4\;{\rm{m/s}}
$

Similarly, the vertical velocity of the projectile is,
${v_x} = \dfrac{{dx}}{{dt}}$
Substitute the expression in the above equation. We get,
$
{v_x} = \dfrac{{d\left( {3t} \right)}}{{dt}}\\
 = 3\;{\rm{m/s}}
$
The velocity of the of a projectile on a certain planet is,
$v = \sqrt {v_x^2 + v_y^2} $
Substitute the values in the above equation. We get,
\[
v = \sqrt {{3^2} + {4^2}} \\
 = \sqrt {9 + 16} \\
 = 5\;{\rm{m/s}}
\]

We have to find the speed of the projection and we know the magnitude of the velocity is equal to the speed of the projectile. So,
$s = 5\;{\rm{m/s}}$

Thus, the speed of projection is 5 m/s.

Note: Velocity is a vector quantity, it has both direction and magnitude while speed is a scalar quantity, and it has only magnitude. Velocity of a body can be equal to zero while the speed of the body cannot be equal to zero.