A projectile is fired with a certain velocity (u) making an angle (theta) with the horizontal.Find the maximum height attained .Total time of flight and horizontal range. Show that it trajectory is parabolic.
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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.
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.
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