Question

a projectile is fired with velocity U making an angle Q with the horizontal . derive expression for maximum height and range​

Answer 1

Suppose it makes an angle=€ with the horizontal.

Then considering the motion along y direction we have,

v(y)=u(y)-gt.

where t is the total time required to attain the max.height.

Also, v(y)=0 at the max. height.

Then, usin€=gt=>t=usin€/g.

Now , Let h(max) be the height. Then,

h(max)=u²×sin²€/2g.

Total time of flight=2usin€/g s.

Then R=ucos€×t=u²×sin2€/g.

Suppose the particle covers y vertical distance in t sec.

Then, y=usin€×t-1/2(g×t²)

=usin€(x/ucos€)-1/2(g)×(x/ucos€)²

where t=x/(ucos€) and x us the horizontal distance covered.