Question

Hi friends ,
here is your question ,


What is projectile motion ?
Derive equations related to it .

Answer 1

an object which is thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible). A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.


Let a projectile be thrown up with a velocity u at angle angle w. Velocity of projectile along the horizontal is u∗cos(w)u∗cos(w) and velocity along the vertical is u∗sin(w).u∗sin(w).

During the motion of the projectile there is no force acting in the horizontal direction. There is only the force of gravity acting in the vertically downward direction. So horizontal component of velocity remains same. Only the vertical component changes. Also the net displacement of the particle is 0.

s=ut+(gt2)/2s=ut+(gt2)/2

We use this equation in the vector form

0=u∗sin(w)∗t−(gt2)/2u∗sin(w)∗t−(gt2)/2

Solving this you get t=2u∗sin(w)/gt=2u∗sin(w)/g

This is the time of flight of the projectile.

During this time the horizontal range covered is R=u∗cos(w)∗2u∗sin(w)/gR=u∗cos(w)∗2u∗sin(w)/g

Simplifying this you get R=(u2)∗sin(2w)/gR=(u2)∗sin(2w)/g

At the highest point of the projectile the vertical component of the velocity becomes 0.

So now we use the formula v2=u2−2asv2=u2−2as

0=(u∗sin(w))2−2gh0=(u∗sin(w))2−2gh

Solving this you get h=((u∗sin(w))2)/2gh=((u∗sin(w))2)/2g

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Answer 2

Here is your answer

Projectile motion is a form of motion experienced by an object or particle (a projectile) that is thrown near the Earth's surface and moves along a curved path under the action of gravity only (in particular, the effects of air resistance are assumed to be negligible).

here we derive equations related to Projectile motion.

Let ,

a projectile be thrown up with a velocity u at angle angle w. Velocity of projectile along the horizontal

u×cos(w)u×cos(w)  and velocity along the vertical is u×sin(w).u×sin(w).

During the motion of the projectile there is no force acting in the horizontal direction. There is only the force of gravity acting in the vertically downward direction. So horizontal component of velocity remains same. Only the vertical component changes. Also the net displacement of the particle is 0.

s=ut+(gt2)/2s=ut+(gt2)/2

We use this equation in the vector form

0=u×sin(w)×t−(gt2)/2u×sin(w)×t−(gt2)/2

Solving this you get t=2u×sin(w)/gt=2u×sin(w)/g

This is the time of flight of the projectile.

During this time the horizontal range covered is 

R=u×cos(w)×2u×sin(w)/gR=u×cos(w)×2u×sin(w)/g

Simplifying this you get :-

R=(u2)×sin(2w)/gR=(u2)×sin(2w)/g

At the highest point of the projectile the vertical component of the velocity becomes 0.

Now

we use the formula.

v2=u2−2asv2=u2−2as

0=(u×sin(w))2−2gh0=(u×sin(w))2−2gh

Hence

 h=(u×sin(w)2)/2g