If an object is dropped from a tall building and hits the ground 3.0 s. later, what is the magnitude of the object's displacement in 1.0 s.; in 2.0 s.?
for 1.0s i got 9.8
for 2.0s i got 20.
is this correct?
doesn't matter how long it takes to hit the ground.
s = 1/2 at^2 = 4.9t^2
s(1) = 4.9
s(2) = 19.6
How did you get 9.8 for s(1)? Curious, since you got s(2) correct. Ah. you must have used 9.8t, rather than 4.9t^2.
The speed is 9.8t. Better review those formulas some more.
To find the magnitude of an object's displacement at different time intervals, we need to consider the object's motion under free fall. Let's break down the problem step by step.
First, we know that an object falls under the influence of gravity, which causes it to accelerate downward at a constant rate of 9.8 m/s² (assuming no air resistance). This acceleration is often denoted by "g."
Now, let's calculate the object's displacement at different time intervals:
1. For t = 1.0 s:
We can use the equation for displacement under constant acceleration:
displacement = initial velocity * time + (1/2) * acceleration * time²
Since the initial velocity is zero (the object starts from rest), and the acceleration is -9.8 m/s² (taking downward as the negative direction), we get:
displacement = 0 * 1 + (1/2) * (-9.8) * (1)²
displacement = -4.9 m (Note the negative sign indicates downward motion.)
So, the magnitude of the object's displacement at t = 1.0 s is 4.9 m.
2. For t = 2.0 s:
Similar to the previous calculation, the equation becomes:
displacement = 0 * 2 + (1/2) * (-9.8) * (2)²
displacement = -19.6 m (Again, negative sign indicates downward motion.)
Therefore, the magnitude of the object's displacement at t = 2.0 s is 19.6 m.
Hence, the correct answers are:
- Magnitude of displacement at 1.0 s: 4.9 m
- Magnitude of displacement at 2.0 s: 19.6 m