Kinematics in two dimension problem:
a baseball pitcher attempts to hit a falling bag of money by throwing a ball at it. the pitcher throws the ball with an initial velocity of v° (v knot) at an angle of ∅ above the horizontal at the same instant that the bag is dropped some distance away(see diagram). three seconds later, the ball hints the bag! what was the value of v° (v knot) and ∅ ?

∅ = 36.9°, v = 25m/s

Explanation:

Since the ball and the bag are affected by gravity in the same way, the pitcher has to aim the ball directly at the bag and the angle can be calculated by the tan∅ = 45m/60m.

The x-component of the velocity v is constant and can be calculated by:

vₓ= 60m/ 3s = 20m/s

The velocity v can be found from this equation:

v = vₓ/cos∅ = 25m/s

25 m/s, 36.9°

Explanation:

The vertical position of the bag at time t is:

y = y₀ + v₀ t + ½ at²

y = 45 + (0) t + ½ (-9.8) t²

y = 45 − 4.9 t²

The vertical position of the ball at time t is:

y = y₀ + v₀ t + ½ at²

y = 0 + (v₀ sin θ) t + ½ (-9.8) t²

y = (v₀ sin θ) t − 4.9 t²

The two are equal when t = 3.

45 − 4.9 (3)² = (v₀ sin θ) (3) − 4.9 (3)²

15 = v₀ sin θ

The horizontal position of the ball at 3 seconds is 60 m:

x = x₀ + v₀ t + ½ at²

60 = 0 + (v₀ cos θ) (3) + ½ (0) (3)²

20 = v₀ cos θ

Two equations, two unknowns.  First, divide the first equation by the second to find θ.

15/20 = (v₀ sin θ) / (v₀ cos θ)

0.75 = tan θ

θ ≈ 36.9°

Next, use Pythagorean theorem to find v₀.

v₀² = (v₀ cos θ)² + (v₀ sin θ)²

v₀² = 20² + 15²

v₀ = 25 m/s

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