Attempting to beat the record for tallest Lego
structure, a student drops a piece from a height
of 24.91 m. How fast will the piece be travelling
when it is 5.0 m above the ground and how long
will it take to get there?
To find the velocity of the piece when it is 5.0 m above the ground, we can use the principles of projectile motion. Assuming there is no air resistance, we can use the equation:
vf^2 = vi^2 + 2ad
Where vf is the final velocity, vi is the initial velocity, a is the acceleration (in this case, due to gravity, which is approximately -9.8 m/s^2), and d is the displacement.
First, let's find the initial velocity (vi). We know that the piece is dropped from a height of 24.91 m, so at that point, its velocity is 0 m/s (since it's not given that the piece was thrown downwards). Therefore, vi = 0 m/s.
Now, let's find the displacement (d). We're interested in the displacement when the piece is 5.0 m above the ground, so d = 5.0 m.
We can now solve for vf:
vf^2 = vi^2 + 2ad
vf^2 = 0^2 + 2(-9.8)(5.0)
vf^2 = -98
vf ≈ -9.9 m/s
The velocity will be approximately -9.9 m/s when the piece is 5.0 m above the ground. The negative sign indicates that the velocity is directed downwards.
To find the time it takes to reach this point, we can use the equation:
d = vit + (1/2)at^2
Since the initial velocity (vi) is 0, the equation simplifies to:
d = (1/2)at^2
Plugging in the values:
5.0 = (1/2)(-9.8)t^2
t^2 = -5.0 / (1/2)(-9.8)
t^2 = 5.0 / 4.9
t ≈ √(5.0 / 4.9)
t ≈ 1.02 s
Therefore, it will take approximately 1.02 seconds for the piece to reach a height of 5.0 m above the ground.