Question

you throw an apple out a window at a height of 6.0 meters upwards at an angle of 30.° to the horizontal at a velocity of 7.0 m/s. what is the maximum height that the apple will reach? round the answer to the nearest tenth.

Explanation:

Step1: Find the initial vertical velocity

The vertical - component of the initial velocity $v_{0y}$ is given by $v_{0y}=v_0\sin\theta$, where $v_0 = 7.0$ m/s and $\theta = 30^{\circ}$. So, $v_{0y}=7.0\times\sin30^{\circ}=7.0\times\frac{1}{2}=3.5$ m/s.

Step2: Use the kinematic equation for vertical motion to find the additional height $\Delta h$

The kinematic equation $v_y^2 - v_{0y}^2=-2g\Delta h$. At the maximum - height, $v_y = 0$. We can solve for $\Delta h$: $\Delta h=\frac{v_{0y}^2}{2g}$, where $g = 9.8$ m/s². Substituting $v_{0y}=3.5$ m/s and $g = 9.8$ m/s², we get $\Delta h=\frac{3.5^2}{2\times9.8}=\frac{12.25}{19.6}\approx0.625$ m.

Step3: Calculate the maximum height $H$

The initial height $h_0 = 6.0$ m. The maximum height $H=h_0+\Delta h$. Substituting $h_0 = 6.0$ m and $\Delta h = 0.625$ m, we get $H=6.0 + 0.625=6.625\approx6.6$ m.

Answer:

$6.6$