Berry Bonds hits a baseball ball with an initial speed of 35m/s at an angle of 50 degrees with respect to the horizon. The playing field is surrounded by a 4.0m tall wall. Assume that when hit, the ball is initially very close to the ground. If the horizontal distance from the point where the ball is hit to the edge of the field is 117m, the ball will..... a) fly over the wall... b)Hit the wall somewhere between 2m and 4m ......c) Hit the wall above ground but below 2m.... d) Hit the ground before reaching the wall ....
And the reason please.
I used 35sin50 as vertical component and 35cos50 as horizontal component, but had a difficult time figuring it out.
Thank you in advance.
To determine whether the ball will clear the wall, hit the wall between 2m and 4m, hit the wall above the ground but below 2m, or hit the ground before reaching the wall, we can break down the problem into two parts: the vertical motion and the horizontal motion.
First, let's analyze the vertical motion. We can use the kinematic equation to determine the height of the ball at any given time. The equation we'll use is:
h = viy * t + (1/2) * a * t^2,
where h is the height, viy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (approximately -9.8m/s^2).
Since the ball starts at ground level, the initial height is 0. The final height is the height of the wall, which is 4m. We can rearrange the equation to solve for the time it takes for the ball to reach the wall:
4 = (viy * t) + (1/2) * (-9.8) * t^2.
Since we know the angle at which the ball was hit, we can find the initial vertical velocity, viy, by using:
viy = vi * sin(theta),
where vi is the initial speed of 35m/s and theta is the angle of 50 degrees.
So, viy = 35 * sin(50) = 35 * 0.766 = 26.81m/s.
Now, we can substitute the values into the equation:
4 = (26.81 * t) + (1/2) * (-9.8) * t^2.
Simplifying the equation, we have:
4 = 26.81t - 4.9t^2.
Rearranging, we get:
4.9t^2 - 26.81t + 4 = 0.
Solving this quadratic equation, we find two possible values for t, let's call them t1 and t2. One of these values will be positive, representing the time it takes for the ball to reach the height of the wall.
Once we know the time it takes for the ball to reach the height of the wall, we can determine how far the ball traveled horizontally during that time using the horizontal component of the initial velocity:
vix = vi * cos(theta).
vix = 35 * cos(50) = 35 * 0.643 = 22.505m/s.
Since we have the time, we can find the horizontal distance using the equation:
x = vix * t.
If this distance is greater than 117m, the ball will fly over the wall (option a). If it is between 2m and 4m, the ball will hit the wall in that range (option b). If it is above ground but below 2m, the ball will hit the wall below 2m (option c). And if it is less than 117m, the ball will hit the ground before reaching the wall (option d).
It's important to note that this analysis assumes ideal conditions, neglecting air resistance and other factors that might affect the trajectory of the ball.
To determine if the ball will fly over the wall, hit the wall somewhere between 2m and 4m, hit the wall above ground but below 2m, or hit the ground before reaching the wall, we need to calculate the vertical and horizontal distances traveled by the ball.
Given:
Initial speed of the ball (v) = 35 m/s
Launch angle (θ) = 50 degrees
Height of the wall (h) = 4.0 m
Horizontal distance from the point where the ball is hit to the edge of the field (d) = 117 m
We can start by calculating the time it takes for the ball to reach the wall using the horizontal component of the initial velocity.
Horizontal velocity (Vx) = v * cos(θ)
Vx = 35 * cos(50)
Vx ≈ 22.545 m/s
Using the formula for horizontal distance (d = Vx * t), we can rearrange it to solve for time (t).
t = d / Vx
t = 117 / 22.545
t ≈ 5.19 s
Now, we can calculate the maximum height reached by the ball using the vertical component of the initial velocity.
Vertical velocity (Vy) = v * sin(θ)
Vy = 35 * sin(50)
Vy ≈ 26.907 m/s
Using the formula for maximum height (Hmax) reached in projectile motion:
Hmax = (Vy^2 * sin^2(θ)) / (2 * g),
where g is the acceleration due to gravity (9.8 m/s^2),
Hmax = (26.907^2 * sin^2(50)) / (2 * 9.8)
Hmax ≈ 27.84 m
Comparing the maximum height (27.84 m) with the height of the wall (4.0 m), we can determine where the ball will land:
a) If Hmax > h + 4.0 m, the ball will fly over the wall.
b) If Hmax is between h and h + 4.0 m, the ball will hit the wall somewhere between 2m and 4m.
c) If Hmax is between 2.0 m and h, the ball will hit the wall above ground but below 2m.
d) If Hmax ≤ 2.0 m, the ball will hit the ground before reaching the wall.
Calculating the limits for each option:
a) Hmax > 8.0 m
b) 4.0 m < Hmax < 8.0 m
c) 2.0 m < Hmax < 4.0 m
d) Hmax ≤ 2.0 m
Since Hmax ≈ 27.84 m, which is greater than 8.0 m, we can conclude that the ball will fly over the wall (option a).
Therefore, the answer is: a) The ball will fly over the wall.
The reason is that the maximum height obtained by the ball is significantly greater than the height of the wall. Hence, the ball will clear the wall and continue to travel beyond it.