Physics, asked by himanahu337, 8 months ago

Derieve the equation of motion such that an object having a horizontal motion

Answers

Answered by keerthanas87
0

Explanation:

An object is thrown with a velocity v

o

with an angle θ with horizontal x-axis.

The velocity component along X-axis= V

ox

=V

o

cosθ

The velocity component along Y-axis =v

oy

=v

0

sinθ

At time T=0, no displacement along X-axis and Y-axis.

So, x

o

=0,y

0

=0

At time T=t,

Displacement along X-axis= v

ox

.t=v

0

cosθ----- (1)

Displacement along Y-axis=v

oy

.t=(v

0

sinθ)t−(

2

1

)gt

2

−−−−−−−(2)

Time required for maximum height,

At maximum height, the velocity component along Y-axis is zero. ie;

v

y

=0 Let the time required to reach maximum height is t

max

Therefore, the initial velocity for motion along Y-axis

v

y

=v

0

sinθ−gt

max

=>0=v

0

sinθ−gt

max

=>t

max

=(v

0

sinθ)/g------- (5)

Total time of flight,

To reach maximum height, time required is v

o

sinθ/g. It takes equal time to reach back ground.

So, total time of flight, T

max

=

g

2(v

0

sinθ)

------- (6)

Maximum height reach,]

Let the maximum height reached by the object be H

max

When body of projectile reaches the maximum height, then

v

y

2

=(v

0

sinθ)

2

=2gH

max

=>0=(v

0

sinθ)

2

=2gH

max

∴H

max

=

2g

(v

0

sinθ)

2

---------- (7)

Horizontal range,

Let R is the horizontal range by the projected body.

Here using horizontal component of velocity only as effective velocity to transverse horizontal path.

R=(v

0

cosθ).T

max

=

g

(v

0

cosθ).2(v

0

sinθ)

=v

0

2

sin2θ/g

R=

g

v

0

2

sin2θ

-------- (8)

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