10. This time, William Tell is shooting at an apple that hangs on a tree (see the figure below). The apple is a horizontal distance of 20.0 m away and at a height of 4.00 m above the ground, if the arrow is released from a height of 1.00 m above the ground and hits the apple 0.700 s later, what is the arrow's initial velocity. Ans.: \( 29.59 \mathrm{~m} / \mathrm{s}, 15.12^{\circ} \mathrm{N} \) of E

We start by breaking the arrow’s motion into horizontal (x) and vertical (y) components. Step 1. Horizontal Motion Since there is no horizontal acceleration, the horizontal distance traveled is given by   x = (v₀ cosθ) t We know:   x = 20.0 m   t = 0.700 s So,   v₀ cosθ = x/t = 20.0 m / 0.700 s ≈ 28.57 m/s Step 2. Vertical Motion The vertical displacement is given by:   y – y₀ = (v₀ sinθ) t – (½) g t² The arrow is released from a height y₀ = 1.00 m and hits the apple at y = 4.00 m, so its vertical displacement is:   Δy = 4.00 m – 1.00 m = 3.00 m Plug in t = 0.700 s and g = 9.80 m/s²:   3.00 m = (v₀ sinθ)(0.700 s) – 0.5(9.80 m/s²)(0.700 s)² Calculate the term with gravity:   0.5(9.80)(0.700²) = 4.9 × 0.49 ≈ 2.401 m So,   (v₀ sinθ)(0.700 s) = 3.00 m + 2.401 m = 5.401 m Thus,   v₀ sinθ = 5.401 m / 0.700 s ≈ 7.72 m/s Step 3. Determine the Initial Speed The magnitude of the initial velocity is found using the Pythagorean theorem:   v₀ = √[(v₀ cosθ)² + (v₀ sinθ)²]   = √[(28.57 m/s)² + (7.72 m/s)²]   ≈ √[816.33 + 59.61] = √[875.94] ≈ 29.59 m/s Step 4. Find the Launch Angle Using the ratio:   tanθ = (v₀ sinθ) / (v₀ cosθ) = 7.72 / 28.57 ≈ 0.270   θ = arctan(0.270) ≈ 15.12° Thus, the initial velocity is approximately 29.59 m/s, launched at an angle of about 15.12° (North of East). Final Answer:   Initial velocity ≈ 29.59 m/s, launch angle ≈ 15.12° N of E.