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 17.0 m/s. The cliff is h = 24.0 m above a flat, horizontal beach as shown in the figure.
(**I was able to answer parts a & b, but am having issues with c & d**)

(a) What are the coordinates of the initial position of the stone?
x0 = 0m
y0 = 24 m

(b) What are the components of the initial velocity?
v0x = 17.0 m/s
v0y = 0 m/s

(c) How long after being released does the stone strike the beach below the cliff?
(d) With what speed and angle of impact does the stone land?
vf = m/s
? =

Explanation / Answer

vx (speed in horizontal direction) = 17m/s (constant)
uy (initial vertical speed) = 0
s (distance) = 24m
a (grav rate) = 9.8m/s/s

Use s = ut + (at^2)/2 to find t (time taken to fall vertically)

24 = 0 + (9.8t^2)/2
9.8t^2 = 48
t = 2.21s

We know horizontal speed is constant. So find out how fast it will accelerate to vertically in 2.21s using

v^2 = u^2 + 2as
v^2 = 0 + 2 x 9.8 x 24
= 470.4
v (vertical) = 21.68 m/s

Using vertical and horizontal speeds, calculate the vector (a^2 = b^2 + c^2)
v^2 = 21.68^2 + 17^2
= 759.0
v = 27.55m/s is the speed it hits the beach.

Use v(total) = v(horizontal) cos x
to find angle x
27.55 = 17 cos x
cos x = 1.62

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