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A body is projected horizontally from a point above the ground and motion of the body is described by the equation $x = 2t$, $y = 5{t^2}$ where $x$, and $y$ are horizontal and vertical coordinates in meter after time $t$. The initial velocity of the body will be
A) $\sqrt {29} m/s$ horizontal
B) $5m/s$ horizontal
C) $2m/s$ vertical
D) $2m/s$ horizontal

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Last updated date: 23rd Feb 2024
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IVSAT 2024
Answer
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Hint: Here we are given with a projected body and asked with the initial velocity of it. The coordinates of movement in both $x$ and $y$ coordinates are given. So it is worth thinking that initially when the projectile body starts moving it moves in the horizontal direction. So, the velocity would also be horizontal. Now, use the concept of differentiation in order to get the magnitude of velocity and your question will be solved.

Complete step by step solution:
Here it is given in the question that the body is projected horizontally from a point. So, the initial motion the body would be in the horizontal direction.
So, here the horizontal velocity would be given as the initial velocity that is asked in the question.
For calculating the velocity at a given point we have to differentiate distance with time.
So, here taking the horizontal coordinates that is the $x$ - coordinates we have, $2m{s^{ - 1}}$
${v_i} = \dfrac{{dx}}{{dt}}$
Putting $x = 2t$ we have,
${v_i} = \dfrac{{d(2t)}}{{dt}}$
On simplifying we have,
${v_i} = 2\dfrac{{dt}}{{dt}}$
As we know that, $\dfrac{{dt}}{{dt}} = 1$
So we have, ${v_i} = 2m{s^{ - 1}}$
So, the initial velocity is $2m{s^{ - 1}}$.

This initial velocity is in horizontal direction so the correct option here is option D that is $2m{s^{ - 1}}$ horizontal.

Note: It is important to note that the horizontal velocity of a projectile remains constant throughout the journey, this can also be proved by the fact that the differentiated value we got for the velocity is a constant, there is no variable in it. So it would not change with time and will remain constant throughout the journey. The value will remain $2m{s^{ - 1}}$ for the horizontal velocity throughout the journey.