A hoodlum throws a rock straight down from the roof of a building 109m high. If it takes 2.40s to hit the ground below, with what initial speed was the rock thrown?
Please show all work
If thrown down with initial velocity v, then just solve
109 - vt - 4.9t^2 = 0
plug in t=2.40 and find v.
To find the initial speed of the rock, we can use the equation of motion for an object in free fall:
s = ut + (1/2)at^2
where:
s = displacement (109m)
u = initial velocity (unknown)
t = time (2.40s)
a = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
Let's solve the equation for the initial velocity (u):
109 = u(2.40) + (1/2)(-9.8)(2.40)^2
First, let's simplify the equation:
109 = 2.40u - 4.9(2.40)^2
Now, expand the right side of the equation:
109 = 2.40u - 4.9(5.76)
Simplify further:
109 = 2.40u - 28.224
Now, isolate the "u" term:
2.40u = 109 + 28.224
2.40u = 137.224
Finally, solve for u:
u = 137.224 / 2.40
u ≈ 57.18 m/s
Therefore, the initial speed at which the rock was thrown was approximately 57.18 m/s.