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Q10 A hiker throws a stone from the upper edge of a vertical cliff. The stones initial velocity is 25.0 m/s directed at 40.0° with the face of the cliff, as shown in the figure. The stone hits the ground 3.75 s after being thrown and feels no appreciable air resistance as it falls. How far from the foot of the cliff does the stone land? (A) 6 m (B) 25.0 m (C) 93.8 m (D) 60.3 Im (E) 0.00 m 25.0 m/s Cliff / Ground

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Answer #1

here,

the initial speed, u = 25 m/s

theta = 40 degree

the time taken to hit the ground , t = 3.75 s

the horizontal distance of stone from the foot of the cliff , x = u * sin(theta) * t

x = 25 * sin(40) * 3.75 m

x = 60.3 m

the horizontal distance of stone from the foot of the cliff is D) 60.3 m

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