Question

you throw an orange out a window at a height of 12.00 meters upwards at an angle of 32° to the horizontal at a velocity of 3.5 m/s. what is the maximum height that the orange will reach? round the answer to the nearest hundredth. meters

Explanation:

Step1: Find the initial vertical velocity

$v_{0y}=v_0\sin\theta$, where $v_0 = 3.5$ m/s and $\theta = 32^{\circ}$. So $v_{0y}=3.5\times\sin(32^{\circ})\approx3.5\times0.5299 = 1.85465$ m/s.

Step2: Use the kinematic - equation to find the additional height

The kinematic equation $v_y^2 - v_{0y}^2=-2gh$ is used to find the additional height $\Delta h$ the orange travels upwards. At the maximum - height, $v_y = 0$. So, $h=\frac{v_{0y}^2}{2g}$, where $g = 9.8$ m/s². Then $h=\frac{(1.85465)^2}{2\times9.8}\approx\frac{3.439}{19.6}\approx0.17546$ m.

Step3: Calculate the maximum height

The maximum height $H$ is the sum of the initial height $h_0 = 12.00$ m and the additional height $\Delta h$. So $H=h_0 + h=12.00+0.17546\approx12.18$ m.

Answer:

$12.18$