When conducting a free fall experiment,Ali drop an egg from a 46 m tall building,accidently encik Abu is walking alongside the building was hit by the egg on the top of his head. If encik Abu height is 1.80m and he is walking alongside the building with constant speed of 1.20ms-1 ,where encik Abu be when Ali release the egg?
encik = Mr.
how long does it take the egg to fall 46-1.8 = 44.2 m?
4.9t^2 = 44.2
t = 3.00 s
During those 3 seconds, Abu has walked 3*1.2 = 3.6 m
To determine where Encik Abu will be when Ali releases the egg, we need to consider the time it takes for the egg to fall from the top of the building to the ground.
In free fall experiments, the distance fallen by an object can be calculated using the equation:
d = (1/2) * g * t^2
where:
d = distance fallen
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken for the fall
Since the height of the building is given as 46 m, we can rearrange the equation to solve for time:
46 = (1/2) * 9.8 * t^2
By simplifying the equation, we find:
t^2 = (46 * 2) / 9.8
t^2 = 9.38
t ≈ 3.06 seconds
Therefore, it will take approximately 3.06 seconds for the egg to fall from the top of the building to the ground.
Now let's determine how far Encik Abu will have walked in that time.
Encik Abu's walking speed is given as 1.20 m/s, and the time taken for the egg to fall is 3.06 seconds. Using the equation:
distance = speed * time
distance = 1.20 * 3.06
distance ≈ 3.67 meters
This means that when Ali releases the egg, Encik Abu will have walked approximately 3.67 meters away from the starting point.
Therefore, Encik Abu will be approximately 3.67 meters away from the base of the building when Ali releases the egg.