Question

8 formula 1 point 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 1.6 meters from the floor below, calculate the maximum height from the floor that the dart reaches. answer

Explanation:

Step1: Find the initial vertical velocity

The initial velocity $v_0 = 14$ m/s and the angle $\theta=60^{\circ}$. The initial vertical velocity $v_{0y}=v_0\sin\theta$. So $v_{0y}=14\sin60^{\circ}=14\times\frac{\sqrt{3}}{2}=7\sqrt{3}$ m/s.

Step2: Calculate the height the dart rises above the initial - height

We use the kinematic equation $v_y^2 = v_{0y}^2-2gh_1$, at the maximum - height, $v_y = 0$. Rearranging for $h_1$, we get $h_1=\frac{v_{0y}^2}{2g}$. Given $g = 9.8$ m/s², $h_1=\frac{(7\sqrt{3})^2}{2\times9.8}=\frac{147}{19.6}=7.5$ m.

Step3: Calculate the maximum height from the floor

The initial height $h_0 = 1.6$ m. The maximum height $H=h_0 + h_1$. So $H=1.6+7.5 = 9.1$ m.

Answer:

9.1 m