Question #192018

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18m/s.  The cliff is 50.m above a flat horizontal beach.  

How long after being released does the stone strike the beach below the cliff?  

With what velocity does it land?



1
Expert's answer
2021-05-11T16:35:00-0400

(a) We can find the time that the stone takes to reach the beach below the cliff from the kinematic equation:


y=12gt2,y=\dfrac{1}{2}gt^2,t=2yg,t=\sqrt{\dfrac{2y}{g}},t=250 m9.8 ms2=3.19 s.t=\sqrt{\dfrac{2\cdot50\ m}{9.8\ \dfrac{m}{s^2}}}=3.19\ s.

(b) Let's first find the vertical component of the stone's velocity:


vy=gt=9.83.19 s=31.3 ms.v_y=-gt=-9.8\cdot3.19\ s=-31.3\ \dfrac{m}{s}.

The sign minus means that the vertical velocity of the stone directed downward.

Finally, we can find with what velocity the stone lands from the Pythagorean theorem:


v=vx2+vy2,v=\sqrt{v_x^2+v_y^2},v=(18.0 ms)2+(31.3 ms)2=36.1 ms.v=\sqrt{(18.0\ \dfrac{m}{s})^2+(-31.3\ \dfrac{m}{s})^2}=36.1\ \dfrac{m}{s}.

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