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

Question

a student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 14m/s. the cliff is 53 m above a flat, horizontal beach as shown in the figure below.

what are the coordinates of the initial position of the stone?

what are the components of the initial velocity?

write the equations for velocity of the stone with time.

write the equations for the position of the stone with time?

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

with what speed and angle of impact does the stone land?

Explanation / Answer

if we take cliff's ground point as origin , then initial position is (0 , 53 m )

initial velocity = 14i m/s   (horizontally) or ux = 14 m/s and uy = 0 m/s

horizontal component will not change so vx = ux = 14 m/s

vy = at = 9.8t

v = 14i + 9.8t   m/s    and magnitude |v| = (196 + 96.04t2 ) m/s

S = 14t i + gt2 /2 j   m/s

S = 14t i + 4.9t2 j m/s

sy = uy t + ay t2 /2

53 = 0 + g X t2 /2

t = 3.29 sec

putting value of t in equation of v

vf = 14i + 9.8 x 3.29 j = 14i + 32.24j m/s

|vf | = (142 + 32.242 ) = 35.14 m/s

= tan-1 ( 32.24 / 14) = 66.54 degrees with the horizontal

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