Question

chapter 6 motion in two dimensions additional problems lesson 1 projectile motion challenge practice 1. you are visiting a friend from elementary school who now lives in a small town. one local amusement is the ice - cream parlor, where stan, the short - order cook, slides his completed ice - cream sundaes down the counter at a constant speed of 2.0 m/s to the servers. (the counter is kept very well polished for this purpose.) if the servers catch the sundaes 7.0 cm from the edge of the counter, how far do they fall from the edge of the counter to the point where the servers catch them? 2. a rock is thrown from a 50.0 - m - high cliff with an initial velocity of 7.0 m/s at an angle of 53.0° above the horizontal. find its velocity when it hits the ground below.

Explanation:

Step1: Analyze vertical - motion

The vertical - motion of the rock is a free - fall motion. The initial vertical velocity is $v_{0y}=v_0\sin\theta$, where $v_0 = 7.0\ m/s$ and $\theta = 53.0^{\circ}$, so $v_{0y}=7.0\sin53.0^{\circ}\approx7.0\times0.799 = 5.593\ m/s$. The height of the cliff is $y_0 = 50.0\ m$ and the acceleration due to gravity $g = 9.8\ m/s^{2}$. The vertical displacement is $y - y_0=- 50.0\ m$. We use the equation $v_y^{2}=v_{0y}^{2}-2g(y - y_0)$.

Step2: Calculate $v_y$

Substitute the values into the equation: $v_y^{2}=(5.593)^{2}-2\times9.8\times(-50.0)$.
$v_y^{2}=31.28 -(-980)=31.28 + 980=1011.28$.
$v_y=\sqrt{1011.28}\approx31.8\ m/s$ (taking the negative root since the rock is moving downwards).
The initial horizontal velocity is $v_{0x}=v_0\cos\theta=7.0\cos53.0^{\circ}\approx7.0\times0.602 = 4.214\ m/s$. Since there is no acceleration in the horizontal direction ($a_x = 0$), $v_x=v_{0x}=4.214\ m/s$.

Step3: Calculate the final velocity

The final velocity $v$ when the rock hits the ground is given by $v=\sqrt{v_x^{2}+v_y^{2}}$.
$v=\sqrt{(4.214)^{2}+(31.8)^{2}}=\sqrt{17.76 + 1011.24}=\sqrt{1029}\approx32.1\ m/s$.

Answer:

The velocity of the rock when it hits the ground is approximately $32.1\ m/s$.