In Figure 4-44, a baseball is hit at a height h = 1.10 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.1 s after it is hit and then down past the top of the wall 4.8 s later, at distance D = 47 m farther along the wall. (a) What horizontal distance is traveled by the ball from hit to catch? What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is the wall?
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To solve this problem, we will break it down into smaller parts and use physics equations and principles to find the answer.
(a) To find the horizontal distance traveled by the ball from hit to catch, we need to find the time it takes for the ball to travel horizontally. We can use the horizontal motion equation: distance = velocity x time.
The initial horizontal velocity of the ball is the same as its final horizontal velocity because it is caught at the same height. Let's call it v_x.
The time it takes for the ball to reach the top of the wall can be found by using the vertical motion equation: height = initial velocity x time + (1/2) x acceleration x time^2. The initial velocity in the vertical direction is 0 m/s since the ball is being hit horizontally.
Using this equation, we can find the time t1 it takes for the ball to reach the top of the wall.
Similarly, the time it takes for the ball to come back down and pass the top of the wall can be found by using the equation: t2 = t1 + 4.8 s.
The total time it takes for the ball to travel horizontally is the sum of t1 and t2: t_total = t1 + t2.
Now, we can calculate the horizontal distance traveled by the ball using the equation: distance = v_x x t_total.
(b) To find the magnitude of the ball's velocity just after being hit, we need to find its total velocity.
From the given information, we know that the ball travels up past the top of the wall 1.1 s after being hit. Using this time and the equation for vertical motion, we can find the vertical component of velocity (v_y) just after being hit.
The total velocity (v) can be found using the Pythagorean theorem: v^2 = v_x^2 + v_y^2. Taking the square root of both sides of the equation gives us the magnitude of the velocity.
(c) To find the angle of the ball's velocity relative to the horizontal, we can use the equation: angle = arctan(v_y / v_x).
(d) To find the height of the wall, we need to find the maximum height reached by the ball. We can use the equation for vertical motion: height = initial velocity x time + (1/2) x acceleration x time^2. The initial velocity in the vertical direction is 0 m/s, and the acceleration is the acceleration due to gravity (9.8 m/s^2).