Question

Heya Friends !!

What is a projectile . Derive an expression for its horizontal range , time of flight , maximum height and prove that it's trajectory are parabola .

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Answer 1

Heya.....

☆☆A projectile is any object thrown into space (empty or not) by the exertion of a force. Although any object in motion through space (for example a thrown baseball) may be called a projectile.

☆☆Horizontal max distance travelled by a particle is called its range....

Here is its expression

R= horizontal speed × flight time

R= ucos theta × 2uSin theta /g

R= u^2 sin 2 theta /g


☆☆3rd one see in attachment....


☆☆let a projectile moves with u velocity which inclined with horizontal ∅ angle .
then,
velocity vector after t time (V)
V = Vx i + Vy j
V = ucos∅ i + (usin∅ -gt) j

at maximum height velocity of projectile have only x -components exist .e.g
Vx =ucos∅
use , Vy² = Uy² + 2ay.Y
for projectile
Vy at maximum height = 0
Uy = usin∅
ay = -g
Y = maximum height

put this ,
0 = (usin∅)² -2gHmax

Hmax = u²sin²∅/2g


☆☆x motion: There is no force in x direction. Therefore , x motion is motion with uniform velocity. In the present problem ,it ( velocity) is v.

The displacement in x direction in time t is

x=vt…………………………………………………..(1)

Now, y motion: for this motion initial velocity is zero. There is acceleration due to gravity in the negative y direction. Then, displacement in y direction is

y=-(1/2)gt^2………………………………………..(2)

Now, we are interested in trajectory that is relation between x and y. Therefore ,we have to eliminate t from eq.(1) and eq.(2).

From eq.(1), t^2=(x/v)^2

Putting this value of t^2 in eq.(2), we get

y =-(1/2)gx^2/v^2……………………………………(3)

Since, g and v are constant. the equation is of the form

y=-ax^2, which is equation of parabola passing through origin and falling in fourth quadrant .

Hence, trajectory in the present problem is a parabola.


Hope it helps...

@skb