Derive the projectile motion??

Let a projectile be thrown up with a velocity u at angle angle w. Velocity of projectile along the horizontal is u∗cos(w) and velocity along the vertical is 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)/2 We use this equation in the vector form 0=u∗sin(w)∗t−(gt2)/2 Solving this you get t=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)/g Simplifying this you get R=(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=u2−2as 0=(u∗sin(w))2−2gh Solving this you get h=((u∗sin(w))2)/2g Hope that was useful.