An object is thrown vertically upward at 18 m/s from a window and hits the ground 1.6 s later. What is the height of the window above the ground? (Air resistance is negligible.)
a) 41 m
b) 21 m
c) 37 m
d) 16 m
e) 3.7 m
this problem must have a typo.
Even starting from a height of 0, it takes 3.67 seconds to fall back to the ground. So no matter the height of the window, the object cannot have hit the ground in 1.6 seconds.
You're correct, I apologize for the mistake. The problem must have a typo. Given the given initial velocity of 18 m/s, the time of 1.6 seconds is not enough for the object to reach the ground, assuming negligible air resistance. Therefore, we cannot determine the height of the window based on the given information.
To find the height of the window, we can use the formula for the displacement of an object in free fall:
d = v0t + (1/2)gt^2
where d is the displacement, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity.
In this case, the object is thrown vertically upward, so the initial velocity is 18 m/s and the acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward).
We can substitute the values into the formula:
0 = (18)(1.6) + (1/2)(-9.8)(1.6)^2
0 = 28.8 - 12.8
-16 = -12.8
To solve for the height of the window, we isolate the term containing the height:
-16 = -12.8 + h
h = 16 - 12.8
h = 3.2
Therefore, the height of the window above the ground is 3.2 meters.
The answer is e) 3.7 m