A bicyclist on an old bike (combined mass: 92 kg) is rolling down (no pedaling or braking) a hill of height 120 m. Over the cour se of the 384 meters of downhill road, she encounters a constant friction force of 261 Newton. If her speed at the top of the hill is 9 m/s, what is her speed at the bottom of the hill
2 answers:
Answer:
V2 = 15.9m/s
Explanation:
See attachment below.
Answer:
V = 48.49m/s
Explanation:
Given the following information:
Combined mass = 92kg
Hill's height = 120m
Course of ride = 384m
Frictional force = 261N
Initial speed (u) = 9m/s
Final speed (v) = ?
Since we are looking for her speed at the bottom,
Time = distance/speed = 384m/9m.s
Time = 42.67s
we use the equation
H = V²/2g ( equation for maximum heigh of trajectory)
Therefore, plugging the values we have
120 = V²/2×9.8
V² = 9.8×120×2
V = √2352
V = 48.49m/s
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