a ball rolls off the edge of a horizontal at 10m/s, what is the speed of the ball 2s later?
Or if your Physics teacher is picky vertical is 9.8 * 2 = 19.6
To find the speed of the ball 2 seconds later, we need to consider the constant acceleration due to gravity.
Step 1: Identify the known values:
Initial speed (u) = 10 m/s
Time (t) = 2 s
Step 2: Calculate the acceleration (a) due to gravity:
The acceleration due to gravity is a constant value of approximately 9.8 m/s^2. Gravity affects all objects equally, regardless of their mass or size.
Therefore, acceleration (a) = 9.8 m/s^2
Step 3: Calculate the final speed (v) using the formula:
v = u + at
Substituting the known values:
v = 10 m/s + (9.8 m/s^2)(2 s)
Step 4: Calculate the final speed:
v = 10 m/s + 19.6 m/s
Therefore, the speed of the ball 2 seconds later is:
v = 29.6 m/s
To find the speed of the ball 2 seconds later, we can use the equation of motion:
\[v_f = v_i + at\]
Where:
- \(v_f\) is the final velocity,
- \(v_i\) is the initial velocity,
- \(a\) is the acceleration, and
- \(t\) is the time.
In this case, since the ball rolls off the edge without any external forces acting on it horizontally, the acceleration is zero. Therefore, we can simplify the equation to:
\[v_f = v_i\]
The initial velocity of the ball is given as 10 m/s, so the speed of the ball 2 seconds later will also be 10 m/s.
after 2 seconds, its downward speed is 10*2 = 20 m/s
So, you have a velocity with horizontal and vertical components of 10 and 20.
Now you can find the speed.