Question

8 formula 3 points a dart is thrown with an initial velocity of 14 m/s at an angle of 60 degrees from the horizontal. assuming the dart has an initial height of 1.8 meters from the floor below, calculate the maximum height from the floor that the dart reaches. answer previous next

Explanation:

Step1: Find vertical component of velocity

The initial vertical velocity \( v_{0y} = v_0 \sin\theta \), where \( v_0 = 14 \, \text{m/s} \) and \( \theta = 60^\circ \). So \( v_{0y} = 14 \times \sin(60^\circ) = 14 \times \frac{\sqrt{3}}{2} = 7\sqrt{3} \, \text{m/s} \approx 12.124 \, \text{m/s} \).

Step2: Calculate maximum height above initial position

Using the kinematic equation \( v_y^2 = v_{0y}^2 - 2g\Delta y \), at maximum height \( v_y = 0 \). Solving for \( \Delta y \): \( \Delta y=\frac{v_{0y}^2}{2g} \), where \( g = 9.8 \, \text{m/s}^2 \). Substituting values: \( \Delta y=\frac{(12.124)^2}{2\times9.8}\approx\frac{146.99}{19.6}\approx7.5 \, \text{m} \).

Step3: Add initial height to get total maximum height

Initial height \( h_0 = 1.8 \, \text{m} \). Total maximum height \( H = h_0+\Delta y = 1.8 + 7.5 = 9.3 \, \text{m} \) (approximate, more precise calculation: \( \Delta y=\frac{(7\sqrt{3})^2}{2\times9.8}=\frac{147}{19.6}=7.5 \), so \( H = 1.8 + 7.5 = 9.3 \, \text{m} \)).

Answer:

\( 9.3 \, \text{meters} \) (or more precisely, using exact steps: \( \Delta y=\frac{(14\sin60^\circ)^2}{2\times9.8}=\frac{14^2\times\frac{3}{4}}{19.6}=\frac{147}{19.6}=7.5 \), so \( H = 1.8 + 7.5 = 9.3 \, \text{m} \))