Question

A body is projected horizontally from a height of 78.4 m with a velocity 10 ms. Its velocity after 3 seconds is .............. [g = 10 ms] (Take
direction of projection on i and vertically
upward direction on j).

In the above problem angle made by velocity
vector with x axis after 4 seconds is tan-inverse​

Answer 1

Answer:

the answer is 10 icap-30j cap

hope it helps you

Explanation:

v=u+at

v=vx+vy

Answer 2

Answer:

The velocity after 3 seconds is 10 i -30 j.

Explanation:

• A velocity is defined as a speed of an object that is in motion and it indicates the rate of the position.
• It is measured using the standard of the time.

• Initial velocity is defined as the velocity at the interval of the time, t = 0 and it is represented by u.
• It is measured in $ms^{-1}$ that is meter per second.

Given that:

Body projected horizontally from a height of 78.4 m

Velocity =10 ms

Velocity after 3 seconds  will be = g = 10 ms

To find:

Velocity after 3 seconds =?

Solution:

Let us consider that,

Initial horizontal velocity $u_{x}$=10m/s

Initial vertical velocity $u_{y}$=0m/s

Vertical velocity at t=3seconds

Formula to find the velocity:

$v_{y} =u_{y} - gt$

Here,

$v_{y}$ = 0

$u_{y}$ = 10

$gt$= 3

Now, by putting these values, we get,

=0−10×3=−30m/s

=−30j^m/s

Velocity after 3 seconds: v=10i^−30j^

Therefore, the final velocity after 3 seconds will be 10 i -30 j.

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