Physics, asked by shivanyawasnik002, 18 days ago

A projectile is launched with initial velocity
Vo= (2î + 3j) m's. The equation of trajectory of particle is (g = 10 m/s)
a) y = 4x - 6x ^2
b) 4y = 6x - 5x^2
c) 4y = 3x - 10x^2
d)y = 4x - 5x^2​

Answers

Answered by ayushnishad16p6m8n9
7

Answer:

little messy but hope it helps

ans b

Attachments:
Answered by soniatiwari214
1

Concept:

The second equation of motion can be defined as,

s = ut + (1/2) at²

where s is distance, t is time, u is initial velocity and a is the acceleration.

Given:

The velocity of projectile, V= (2î + 3j) m/s

Find:

The equation of trajectory.

Solution:

Given, V= (2i + 3j) m/s

The velocity in the i direction is 2i m/s.

The distance traveled in the i direction, that is in the x-axis, acceleration is zero in the x-axis,

s = ut + (1/2) at²

x = ut + (1/2) (0)t²

x = 2t m/s

The velocity in the j direction is 3j m/s.

The distance traveled in the j direction, that is in the y-axis, acceleration is 10m/s² in the downward y-axis,

s = ut + (1/2) at²

y = (3)t - (1/2) (10)t²

y = 3t - 5t²

So, writing x in terms of y,

y = (3/2)x - (5/4)x²

4y = 6x - 5x²

Hence, the equation of the trajectory is 4y = 6x - 5x²  and the correct option is (b) 4y = 6x - 5x².

#SPJ2

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