Question

a slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. the stone lands 36.0 meters away.
a) how long was the stone falling?
b) at what speed was the stone launched?
c) what is the vertical speed at impact?
d) what is the horizontal speed at impact?
e) at what ccw angle does the stone strike the ground?
a ball rolls with a speed of 2.0 m/s across a level table that is 1.0 m above the floor. upon reaching the edge of the table, it follows a parabolic path to the floor. how far along the floor is the landing spot from the table?

Explanation:

Step1: Calculate time of fall for stone (a)

Use vertical - motion equation $y = v_{0y}t+\frac{1}{2}gt^{2}$. Since $v_{0y} = 0$ (launched horizontally), $y = 20.0$ m and $g = 9.81$ m/s². So, $y=\frac{1}{2}gt^{2}$, and $t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2\times20.0}{9.81}}\approx2.02$ s.

Step2: Calculate launch speed (b)

In horizontal motion, $x = v_{0x}t$. Given $x = 36.0$ m and $t = 2.02$ s. Then $v_{0x}=\frac{x}{t}=\frac{36.0}{2.02}\approx17.82$ m/s.

Step3: Calculate vertical speed at impact (c)

Use $v_{y}=v_{0y}+gt$. Since $v_{0y} = 0$, $v_{y}=gt$. Substituting $g = 9.81$ m/s² and $t = 2.02$ s, we get $v_{y}=9.81\times2.02\approx19.82$ m/s.

Step4: Calculate horizontal speed at impact (d)

In horizontal motion (no - acceleration), $v_{x}=v_{0x}$. So $v_{x}=\frac{x}{t}=\frac{36.0}{2.02}\approx17.82$ m/s.

Step5: Calculate the impact angle (e)

Use $\tan\theta=\frac{v_{y}}{v_{x}}$. Substituting $v_{y}\approx19.82$ m/s and $v_{x}\approx17.82$ m/s, $\theta=\arctan(\frac{19.82}{17.82})\approx48.0^{\circ}$ counter - clockwise.

Step6: Calculate the range for the ball

For the ball rolling off the table, first find the time of fall. Using $y = v_{0y}t+\frac{1}{2}gt^{2}$ with $y = 1.0$ m and $v_{0y}=0$, $t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2\times1.0}{9.81}}\approx0.45$ s. Then, in horizontal motion $x = v_{0x}t$ with $v_{0x}=2.0$ m/s. So $x = 2.0\times0.45 = 0.90$ m.

Answer:

a) $t\approx2.02$ s
b) $v_{0x}\approx17.82$ m/s
c) $v_{y}\approx19.82$ m/s
d) $v_{x}\approx17.82$ m/s
e) $\theta\approx48.0^{\circ}$
For the ball: The landing spot is $0.90$ m from the table.