Kevin tosses a 0.10 kg ball horizontally with an initial velocity of 22.00 m/s. What is in m/s2 the vertical component of the acceleration of the ball 3.82 seconds after he threw it?
To solve this problem, we need to break down the given information and use the basic equations of motion.
First, we know that the ball is tossed horizontally, which means there is no initial vertical velocity. So, the initial velocity in the vertical direction (Vy) is 0 m/s.
Second, we need to find the vertical displacement (Δy) of the ball after 3.82 seconds. We can use the equation:
Δy = Vy * t + (1/2) * a * t^2
Since Vy = 0 m/s, the equation simplifies to:
Δy = (1/2) * a * t^2
Now, let's solve for Δy:
Δy = (1/2) * a * (3.82 s)^2
Next, we need to find the acceleration in the vertical direction (ay). We'll use the equation:
Δy = (1/2) * ay * t^2
We can rearrange the equation to solve for ay:
ay = (2 * Δy) / t^2
Now, substitute the values:
ay = (2 * Δy) / (3.82 s)^2
Given that the mass of the ball is 0.10 kg, we can calculate the vertical displacement (Δy) using kinematic equations. Assuming the ball is in free fall, the vertical displacement can be found using the equation:
Δy = (1/2) * g * t^2
where g is the acceleration due to gravity, approximately 9.8 m/s^2.
Substituting the values, we get:
Δy = (1/2) * 9.8 m/s^2 * (3.82 s)^2
Now calculate Δy.
Finally, substitute the calculated Δy value into the equation for ay:
ay = (2 * Δy) / (3.82 s)^2
Simplify the equation to get the answer.