Question

in figure p3.7. e. how long after being released does the stone strike the beach below the cliff? f. with what speed and angle of impact does the stone land? figure p3.7

Explanation:

Step1: Find time of flight (for part e)

Use vertical - motion equation $y = y_0+v_{0y}t-\frac{1}{2}gt^{2}$. Here, $y = 0$, $y_0=h = 50.0$ m, $v_{0y}=0$ m/s. So, $0 = h+0\times t-\frac{1}{2}gt^{2}$, which simplifies to $t=\sqrt{\frac{2h}{g}}$. Substituting $h = 50.0$ m and $g = 9.8$ m/s², we get $t=\sqrt{\frac{2\times50.0}{9.8}}\approx3.19$ s.

Step2: Find vertical and horizontal components of velocity at impact (for part f)

The horizontal component of velocity remains constant: $v_x=v_{0x}=18.0$ m/s. The vertical component of velocity is given by $v_y = v_{0y}-gt$. Since $v_{0y}=0$ m/s, $v_y=-gt=- 9.8\times3.19\approx - 31.3$ m/s.

Step3: Find the speed at impact

The speed $v$ at impact is given by $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{(18.0)^{2}+(-31.3)^{2}}=\sqrt{324 + 979.69}=\sqrt{1303.69}\approx36.1$ m/s.

Step4: Find the angle of impact

The angle $\theta$ of impact is given by $\tan\theta=\frac{v_y}{v_x}$. So, $\theta=\arctan(\frac{-31.3}{18.0})\approx - 60.0^{\circ}$ (the negative sign indicates the angle is below the horizontal).

Answer:

e. $t\approx3.19$ s
f. Speed $v\approx36.1$ m/s, Angle $\theta\approx - 60.0^{\circ}$