A student stands at the edge of cliff and throwns a stone horizontally over the

Question

A student stands at the edge of cliff and throwns a stone horizontally over the edge with a speed of 18m/s. the cliff is 50.0m above a flat, horizontal beach.
a) what are the coordinates of the initial position of the stone?
b) what are the components of the initial velocity?
c) write the equation for the x- and y-components.
d)how long after being released does the stone strike the beach below the cliff?
e) with what speed and angle of impact does the stone land?
please show all work step by step.

Explanation / Answer

a) x=0,y=50 m

b) Vo=18 m/s==>vox=vx=Vo*cos()=18 m/s==>voy=Vo*sin()=0 m/s

c) x(t)=X0+Vo*t+1/2*ax*t2==>ax=0 because there is acceleration in

the x direction==>y(t)=Y0+Vo*t+1/2*ay*t2==>

d) (1/2*4*9.8*50)/9.83.2 s total flight time

e) vx=18 m/s, vyfinal=voy+g*t=0+9.8*3.2-31.3 m/s==>vfinal=(182+31.32)36.111 m/s

tan-1(-31.3/18) -60.1o

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