Answer:
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
Explanation:
First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:
Vf = Vi + gt
where,
Vf = Velocity of egg at the time of hitting head of the Professor
Vi = initial velocity of egg = 0 m/s (Since, egg is initially at rest)
g = acceleration due to gravity
t = time taken by egg to come down
Therefore,
Vf = 0 + gt
t = Vf/g ----- equation (1)
Now, we use 3rd euation of motion for Vf:
2gs = Vf² - Vi²
where,
s = height dropped = H - h
Therefore,
2g(H - h) = Vf²
Vf = √[2g(H - h)]
Therefore, equation (1) becomes:
t = √[2g(H - h)]/g
t = √[2(H - h)/g]
Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:
s = vt
<u>s = v√[2(H - h)/g]</u>
<u>This formula describes the distance from the building that Prof Murthy must be when you release the egg</u>
