Answer:
23.75 seconds
Explanation:
A sketch of the scenario is attached to this response.
This is a motion of free fall and some of the equations of such motion will be used here.
To solve this problem, follow these steps
(i) First get the height of the building.
The ball falls from rest at the top of a building. It falls 0.3 of the height of the building in 13 seconds...
During free fall, the height (h) of an object is given as;
h = ut + [tex]\frac{1}{2} at^2[/tex] -------------------(i)
Where;
h = height of the object from the reference point to a point in time.
t = point in time
a = acceleration due to gravity = 10m/s² [taking downward motion positive]
u = initial velocity of the object.
In this case;
Let the height of the building be H.
At t = 13 seconds, h = 0.3H
Also, since the object falls from rest, the initial velocity is zero (0)
Substitute these values into equation(i) as follows;
0.3H = (0)t + [tex]\frac{1}{2} (10)(13)^2[/tex]
0.3H = 0 + [tex]\frac{1}{2} (10)(13)^2[/tex]
0.3H = [tex]\frac{1}{2} (10)(13)^2[/tex]
0.3H = 845
H = [tex]\frac{845}{0.3}[/tex]
H = 2816.7m
(ii) Now get the total time of the object's fall in seconds
The total time of its fall is the time taken for it to fall from the top to the ground.
Using the same equation(i) used above;
Let the total time taken to fall from the top to the ground be T
The height from the top to the ground is H = 2816.7m
The initial velocity is 0
Acceleration due to gravity is 10m/s²
Substitute these values into equation(i)
2816.7 = 0 + [tex]\frac{1}{2} (10)T^2[/tex]
2816.7 = [tex]\frac{1}{2} (10)T^2[/tex]
2816.7 = [tex]5T^2[/tex]
T² = [tex]\frac{2816.7}{5}[/tex]
T² = 563.34
T = √563.34
T = 23.75seconds
Therefore, the total time of its fall in seconds is 23.75
