You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 tall, is walking along side the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor's head, how far from the building should the professor be when you release the egg? Assume that the egg is in free fall.Take the free fall acceleration to be&nbsp;9.80&nbsp;ms2

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You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 tall, is walking along side the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor&#039;s head, how far from the building should the professor be when you release the egg? Assume that the egg is in free fall.Take the free fall acceleration to be&amp;nbsp;9.80&amp;nbsp;ms2

Bertha Stark · Accepted Answer

Height of building (h) = 46.0 m&amp;nbsp;Height of walling person (h1) = 1.2&amp;nbsp;ms&amp;nbsp;The egg&#039;s initial speed is (v0x) =0&amp;nbsp;ms&amp;nbsp;Speed of egg&amp;nbsp;(v)=2gh, where&amp;nbsp;g=9.8ms2&amp;nbsp;=2(9.8)(46)=30.03&amp;nbsp;ms&amp;nbsp;Let the egg fall on the people.

Jeffrey Jordon · Accepted Answer

S=μt+12gt2  ⇒44.2=0t+12×98×t2  ⇒t2=3s  ∴ Professor should be at distance of 3×1.2−3.6 m

RizerMix · Accepted Answer

To solve the problem, we can use the equation of motion for an object in free fall:h=12gt2where h is the height, g is the acceleration due to gravity, and t is the time taken.In this case, the height h is the distance between the roof and the ground, which is given as 46.0 m. The acceleration due to gravity g is 9.80 m/s^2.We want to find the time it takes for the egg to fall from the roof to the ground. We can rearrange the equation to solve for t:t=2hgSubstituting the values into the equation:t=2×46.09.80

Vasquez · Accepted Answer

The time it takes for the egg to fall from the roof to the ground can be found using the equation:h=12gt2where h is the height of the building (46.0 m) and g is the acceleration due to gravity (9.80 m/s2). Solving for t, we get:t=2hgNow, during this time t, the professor moves a distance of v·t, where v is the velocity of the professor (1.20 m/s). Therefore, the distance d can be expressed as:d=v·t=v·2hgSubstituting the given values, we have:d=1.20·2·46.09.80≈9.35&amp;nbsp;mTherefore, the professor should be approximately 9.35 meters away from the building when you release the egg in order to hit their head.

nick1337 · Accepted Answer

Step 1:Let&#039;s define the following variables:h = height of the building (46.0 m)v0 = initial velocity of the egg (0 m/s, as it is dropped)a = acceleration due to gravity (−9.80m/s2, as it acts downwards)t = time taken for the egg to fallUsing the kinematic equation:h=v0t+12at2Step 2:We can plug in the known values and solve for t:46.0=0·t+12·(−9.80)·t2Simplifying the equation:46.0=−4.90t2Dividing both sides by -4.90:t2=−46.04.90Taking the square root of both sides:t=−46.04.90Since time cannot be negative in this context, we know that the egg will take the positive square root of the right-hand side.Step 3:Now, we can determine the horizontal distance traveled by the professor during this time. The professor walks at a constant speed of 1.20 m/s, so the distance traveled is given by:d=speed·timeSubstituting the values:d=1.20·−46.04.90To find the magnitude of this distance, we can take the absolute value of the expression:d=|1.20·−46.04.90|Therefore, the professor should be approximately 1.20·−46.04.90 meters away from the building when you release the egg.

You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 tall, is walking along side the building at

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