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Class 10 Class 12

A ball is projected horizontally with a velocity 4g m/s from height 18 meter. What is the velocity of body at t = 3 sec?

Since the body is projected in horizontal direction with velocity 4g, therefore

Now the horizontal and vertical component of velocity at t = 3s is,
Since the body is projected in horizontal direction with velocity 4g,

The angle that the velocity vector makes with horizontal is,
Since the body is projected in horizontal direction with velocity 4g,

129 Views

A body is projected with velocity 19·6 m/s in horizontal direction. After what time the horizontal and vertical component of velocity will be the same?

Given, that the body is projected in the horizontal direction.

So, we have

Now, the horizontal component of velocity at any instant is given by,

Vertical component of velocity is,

Since, horizontal and vertical component of velocity is the same, we have

9.8 t = 19.6

i.e., t = 2s

113 Views

A projectile is projected horizontally with velocity u and allowed to fall only under gravity. Discuss the trajectory of projectile.

Consider a projectile, which is projected with velocity 'u' in horizontal direction from a point at height h from the ground.

Let origin be the point of projection.

The right direction is positive direction of X-axis and vertically downward direction is positive direction of Y-axis.

The component of velocity and acceleration is resolved into horizontal and vertical components.

So,

Motion along horizontal direction is,

...(1)

Motion along vertical direction is,

...(2)

Eliminating 't' from (1) and (2), we get

This is the required equation of trajectory.

Since the equation of trajectory is an equation of parabola, therefore path of projectile is parabola.

132 Views

A projectile is fired with velocity u horizontally from height h. With what velocity will it strike the ground?

Let a projectile be projected with velocity u horizontally from height h. Let the projectile strike the ground with velocity v after time t. Break the components of initial velocity and acceleration along X-axis and Y-axis.

Time taken by the body to reach the ground:

Velocity of the body at ground:
Let a projectile be projected with velocity u horizontally from heigh

185 Views

A projectile is projected horizontally with velocity from the top of an inclined plane of inclination θ. How far from the point of projection will the projectile hit tht plane?

Let us consider a projectile is projected with velocity u in horizontal direction from a point O on an inclined plane and hit the inclined plane at P↔(x, y). Let O be the origin, right direction is positive direction of X-axis and vertically downward direction is positive direction of Y-axis. Break the components of velocity and acceleration due to gravity in horizontal and vertical direction.
Let us consider a projectile is projected with velocity u in horizont

space space space space straight x subscript straight o equals 0 space space space space space space space space straight y subscript straight o equals 0 space space space space straight u subscript straight x equals straight u space space space space space space space space straight u subscript straight y space equals 0 space space space space space straight a subscript straight x equals 0 space space space space space space space space space straight a subscript straight y space equals space straight g

Motion along horizontal direction is,

space straight x equals straight x subscript straight o plus straight u subscript straight x straight t plus 1 half straight a subscript straight x straight t squared equals ut

...(1)
Motion along vertical direction is,

straight y equals straight y subscript straight o plus straight u subscript straight y straight t plus 1 half straight a subscript straight y straight t squared equals 1 half gt squared

...(2)
Eliminating t from (1) and (2),

straight y equals fraction numerator straight g over denominator 2 straight u squared end fraction straight x squared

...(3)
The equation of inclined plane is,

space straight y equals xtanθ

...(4)
The point P is the intersection of trajectory and inclined plane, thus the coordinates of P are the solutions of equation (3) and (4). By solving (3) and (4), we obtain