A woman throws a ball at a vertical wall d = 5.8 m away. The ball is h = 2.9 m above ground when it leaves the woman's hand with an initial velocity of 13 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.)
So much for the facts. What is your question?
To solve this problem, we can break it down into two separate motions: the horizontal motion and the vertical motion.
1. Horizontal motion:
We are given that the ball is thrown at an initial velocity of 13 m/s at an angle of 45 degrees. The horizontal component of the velocity remains constant throughout the motion, so we can find the time it takes for the ball to reach the wall using the horizontal distance and velocity component.
Using the formula: distance = velocity * time, we can rearrange it to solve for time:
time = distance / velocity
Given that the horizontal distance (d) is 5.8 m and the horizontal component of the velocity (Vx) is 13 m/s, we can substitute these values into the formula to find the time it takes for the ball to reach the wall.
time = 5.8 m / 13 m/s
time ≈ 0.446 seconds
2. Vertical motion:
The vertical motion of the ball is influenced by gravity. The ball leaves the woman's hand at a height of 2.9 m above the ground, and the vertical component of its velocity remains unchanged throughout the motion.
We can use the formula: height = initial velocity * time + (1/2) * acceleration * time^2, where the acceleration is -9.8 m/s^2 (negative due to gravity).
Substituting the given values, we have:
2.9 m = 0 + (1/2) * (-9.8 m/s^2) * (0.446 seconds)^2
Simplifying the equation will give us the time it takes for the ball to reach its maximum height.
2.9 m = -(1/2) * (9.8 m/s^2) * (0.199 seconds)^2
2.9 m ≈ -0.8745 m
The height is negative, indicating that the ball does not reach a maximum height at this time.
Now, let's find the total time it takes for the ball to hit the wall by adding the time it took to reach the wall horizontally (0.446 seconds) and the time it took for the ball to be at the same vertical position as the wall (0.199 seconds).
Total time = 0.446 seconds + 0.199 seconds
Total time ≈ 0.645 seconds
Therefore, it takes approximately 0.645 seconds for the ball to hit the wall after it was thrown.