A ball falling from rest is located 45 m below its starting point 3.0 s later. Assuming that its acceleration is uniform, what is its value?
X = (1/2) a t^2
t is the time spent falling, and a is the acceleration. Solve for a using X = 45 m and t = 3.0 s.
a = 2X/t^2 = 90 m/9s^2 = 10 m/s^2
Well, this ball seems to have fallen pretty fast! I would say it must have been in a hurry to get to the ground. Anyway, let's calculate its acceleration.
We know that the ball fell 45 m in 3.0 seconds. So, we can use the formula:
d = (1/2) * a * t^2
Where d is the distance, a is the acceleration, and t is the time.
Now, plugging in the values, we have:
45 m = (1/2) * a * (3 s)^2
Simplifying that equation, we get:
45 m = (1/2) * a * 9 s^2
Now, let's solve for a. We'll start by getting rid of that pesky 1/2:
90 m = a * 9 s^2
Dividing both sides of the equation by 9 s^2, we find:
a = 10 m/s^2
So, the value of the acceleration is 10 m/s^2. It fell pretty fast, didn't it? I hope it didn't make too much of a mess on the way down!
To find the acceleration of the ball, we can use the formula for motion under constant acceleration:
\[ d = ut + \frac{1}{2} a t^2 \]
Where:
d = displacement
u = initial velocity
t = time
a = acceleration
In this case, the ball is falling from rest, so its initial velocity u is 0 m/s. The displacement d is 45 m, and the time t is 3.0 s.
Therefore, the formula becomes:
\[ 45 = 0 \times 3 + \frac{1}{2} a \times 3^2 \]
Simplifying this equation, we get:
\[ 45 = \frac{1}{2} \times 9a \]
\[ 90 = 9a \]
Dividing both sides of the equation by 9, we find:
\[ a = \frac{90}{9} \]
\[ a = 10 \, \text{m/s}^2 \]
Therefore, the acceleration of the ball is 10 m/s^2.
To find the acceleration of the ball, we can use the equation of motion for an object under constant acceleration:
\[d = ut + \frac{1}{2}at^2\]
Where:
- d is the displacement or distance traveled by the object (45 m)
- u is the initial velocity (0 m/s, since the ball is falling from rest)
- a is the acceleration (unknown)
- t is the time taken (3.0 s)
Rearranging the equation, we get:
\[2d = 2ut + at^2\]
Since the initial velocity is 0 m/s, we can simplify the equation further:
\[2d = at^2\]
Substituting the known values, we have:
\[2(45 \, \text{m}) = a(3.0 \, \text{s})^2\]
Simplifying, we get:
\[90 \, \text{m} = 9a\]
Dividing both sides by 9, we find:
\[10 \, \text{m/s}^2 = a\]
Therefore, the acceleration of the ball is 10 m/s^2.