Given:
Initial velocity (u) = 37 m/s (upwards)
Height of the building (h) = 56 m
We'll consider the motion in two parts: the upward motion and the downward motion.
Upward Motion:
Using the equation for displacement in vertical motion:
s = ut + (1/2)at²
At the top of the building, the displacement (s) will be equal to the height of the building (h), which is 56 m.
The initial velocity (u) is 37 m/s (upwards), and the acceleration (a) is the acceleration due to gravity, approximately -9.8 m/s² (negative because it acts downwards).
56 = 37t + (1/2)(-9.8)t²
56 = 37t - 4.9t²
Simplifying the equation and rearranging:
4.9t² - 37t + 56 = 0
Now we can solve this quadratic equation to find the time (t) when the arrow passes the top of the building on its way up.
Downward Motion:
The arrow will take the same time to come down from the top of the building to the ground, neglecting air resistance.
Therefore, the time at which the arrow passes the top of the building on its way up and down will be the same.
By solving the quadratic equation, we can determine the value of t. Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
For the equation 4.9t² - 37t + 56 = 0, substituting the values of a, b, and c:
t = (-(-37) ± √((-37)² - 4(4.9)(56))) / (2(4.9))
Simplifying further:
t = (37 ± √(1369 - 1097.6)) / 9.8
t = (37 ± √(271.4)) / 9.8
Calculating the two possible values of t:
t ≈ (37 + 16.47) / 9.8 ≈ 5.57 seconds (approximately)
t ≈ (37 - 16.47) / 9.8 ≈ 2.08 seconds (approximately)
Therefore, the arrow passes the top of the building on its way up and down at approximately 2.08 seconds and 5.57 seconds.