A stone of mass100 g is thrown vertically upwards with an initial velocity 20 m/s. if 40 % of the initial energy is lost against air friction, what is the maximum height reached by stone??
Answers
Answer:
Initial velocity u = 40m/s
g = 10 m/s2
Max height final velocity = 0
Consider third equation of motion
v2 = u2 – 2gs [negative as the object goes up]
0 = (40)2 – 2 x 10 x s
s = (40 x 40) / 20
Maximum height s = 80m
Total Distance = s + s = 80 + 80
Total Distance = 160m
Total displacement = 0 (The first point is the same as the last point)
Explanation:
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Answer: Max height Reached = 12 m.
Explanation:
According to the law of conservation of energy,
KE (at any point) + PE (at any point) = Total Mechanical Energy __(A)
As 40% of initial KE is lost. Then effective KE is:
KE′ = KE - 40% KE = ½ mv² - ⅖•½mv²
=> KE′ = 3/10 mv²
=> KE′ = 3/10 × 1/10 kg × 20 m/s × 20 m/s
=> KE′ = 12 J (at bottom)
As per (A),
KE at bottom = PE at top
=> PE = 12 J
=> mgh = 12 J
=> 1/10 kg × 10 m/s² h = 12 J
=> h = 12 m