Question

a shell thrown towards a target from the ground with a fixed velocity falls 'p' metres beyond the target when fired at an angel 'a' & falls 'q' meters short when fired at an angle 'b' . show that the correct angle of projection is 1/2xsin^-1 ((p.sin2a +q.sin2b)/(p+q))

Answer 1

Just to make sure: the correct angle theta satisfies
\[\frac{1}{\theta} = 2.\arcsin \left( \frac{p.\sin 2a + q.\sin 2b}{p + q} \right)\]
?

Nice problem.

Answer 2

No ... it is that theta equals 1/2 times the arcsin expression

Answer 3

no
2x\[\theta =\]sin^-1 ((p.sin2a +q.sin2b)/(p+q))

Answer 4

yes. This actually isn't so bad.

Suppose a projectile is fired with velocity v at an angle alpha to the horizontal. What is the expression for how far the projectile travels along the horizontal?

Answer 5

\[R= u ^{2}\sin2\alpha/ g\]

Answer 6

The horizontal velocity is v.cos(alpha) and hence it travels
v.cos(alpha).T
where T is the time the projectile reaches the horizontal line again.

We can find T from considering the vertical velocity, call that vy(t)

vy(t) = v.sin(alpha) - gt

hence vertical distance, dy(t) by integrating and setting dy(0) = 0

dy(t) = v.sin(alpha)t - gt^2/2

We now find T, as it satisfies dy(T) = 0.

So do that and find the expression for the distance travelled by the projectile, call it d(alpha)

Answer 7

Right.

Answer 8

Now you have
d(a) = x - p and d(b) = x + q
where x is the distance we want.

Now manipulate these equations and find the theta such that

d(theta) = x.