33. A cannon ball is shot from the ground with velocity v at an angle alpha (Fig

Question

33. A cannon ball is shot from the ground with velocity v at an angle alpha (Figure 32) so that it has a vertical component of velocity v sin alpha and a horizontal component v cos alpha. Its distance s(t) above the ground obeys the law s(t) = -16t^2 + (v sin alpha)t, while its horizontal velocity remains constantly v cos alpha. (a) Show that the path of the cannon ball is a parabola (find the each time t, and show that these points lie on a parabola). (b) Find the angle alpha which will maximize the horizontal distance traveled by the cannon ball before striking the ground.

Explanation / Answer

Supposing a coordinate plane with origin at the point of throw of ball as shown in fig 32 .Y-axis vertical uward and X-axis horizontal rightwards

Then Vx = Vcos(a)

Vy = Vsin(a)

Sx = V cos(a) *t

t = [Sx / V cos(a) ] ...............(1)

Sy = V sin(a) *t - 1/2gt^2 = V sin(a) * t - 16t^2 ...............(2)

Putting t from (1) in (2)

Sy = Vsin(a)/V cos(a) *Sx - 16 *[Sx/Vcos(a)]^2 = Sx * tan(a) - (16 /V^2 cos^2(a) ) *Sx^2 = C1 Sx - C2 Sx^2

So as [Sy =C1 Sx - C2 Sx^2 ] is the standard form of parabola & eqn(1) & eqn(2) gives position of ball at every t (ans)

b) For horizontal distance, we need to find time of flight

So from (2)

Sy = V sin(a) *t - 16t^2 =0

tf = V sin(a) /16

So R =V cos(a)* tf = V^2 sin(a) *cos(a) /16 = V^2 *sin(2a)/32

So for Rmax =sin(2a) should be maxm which is maxm for the angle a=45 degree (ans)

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