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# Problem on Two and Three dimensional motion - projectile motion?

I had asked this question earlier and the answer I got was "use standard formulas for projectile motion"... obviously I know this that is why I captioned it as "Problem on Two and Three dimensional motion - projectile motion". I also have already tried applying those formulas.

Can anyone help me in this please? Here goes the question -

Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water. Although the fish sees the insect along a straight line path at and angle "phi" and distance "d", a drop must be launched at a different angle "theta 0" if its parabolic path is to intersect the insect. If "phi" = 36 degrees "d" = 0.9 m and the launch speed is 3.56 m/s what "theta 0" is required for the drop to be at the top of the parabolic path when it reaches the insect?

### 1 Answer

- JullyWumLv 79 years agoFavorite Answer
It's a 2D problem.

Assuming the angle θ is measured to the horizontal ..

The insect is at the mid-point of the trajectory, having ..

• a vertical displacement .. h = 0.90.sin36 .. .. h = 0.53m

• a horiz. displacement .. x = 0.90.cos36 .. .. . x = 0.73m

The range .. R = 2x = (u².sin 2θ / g)

x = (u².sin 2θ / 2g) ------------ (1)

Applying .. [ Vv² = Uv² - 2gh ] to the drop's vertical motion .. .. (Vv = 0 at displacement h at top of trajectory)

Uv² = 2gh

Since Uv = u.sinθ .. .. u² = 2gh / (sin θ)² ------------ (2)

Substitute (2) into (1)

x = [2gh / sin²θ].sin 2θ/2g .. .. .. x = h.sin2θ / sin²θ

Since .. sin2θ = 2.sinθ.cosθ .. .. x = h.2.cosθ / sinθ .. .. x = 2h / tan θ

Tan θ = 2h/x .. .. 2 x 0.53m / 0.73m .. .. tanθ = 1.4521 .. .. .. ►θ = 55.0º