Question

As shown in figure a stone is projected at a cliff of helght h with an
initial speed of 42.0 mis directed at angle 0,= 60.0° above horizontal,
The stone strikes al 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. (g = 10 m/s)
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Answer 1

Answer:

Step-by-step explanation:

initial Speed = 42 m/s

Horizontal speed = 42Cos60 = 21 m/s

Vertical Speed = 42Sin60 = 21√3 m/s

Horizontal Distance in 5.5 Sec = 21 * 5.5 = 115.5 m

Velocity after  5.5 Sec = 21√3  - g*5.5

= 36.33 - 55

= -18.67 m/s

18.67 m/s  downward

MAx height = -(21√3)²/(2* (-10)) = 66.15 m

Height of cliff = 21√3*5.5 - (1/2)(10)(5.5)²

= 199.82 - 151.25

= 48.57 m