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Question

In Fig. 4-34, a stone is projected at a cliff of height h with an initial speed of 42.0m/s directed at angle theta0=60.0° above the horizontal. The stone strikes at A, 5.50s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.
1769739_650d7316fba14847b46bf159857fc7ca.png

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Solution

(a) Using the same coordinate system assumed in Eq. , we solve for y=h:
h=y0+v0sinθ0t12gt2
which yields h=51.8m for y0=0, v0=42.0m/s, θ0=60.0°, and t=5.50s.

(b) The horizontal motion is steady, sovx=v0x=v0cosθ0 but the vertical component of
velocity varies according to Eq. . Thus, the speed at impact is
v=(v0cos0)2+(v0sinθ0gt)2=27.4m/s

(c) We use Eq. with vy and y=H:
H=(v0sinθ0)22g=67.5m.

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