A ball player hits a home run, and the baseball just clears a wall 17.0 m high located 127.0 m from home plate. The ball is hit an an angle of 35° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.0 m above the ground.
(a) What is the initial speed?
m/s
(b) How much time does it take for the ball to reach the wall?
s
(c) Find the components of the velocity and the speed of the ball when it reaches the wall
vy,f = m/s
vx,f = m/s
vf = m/s
To solve this problem, we can break it down into multiple steps. Let's start step by step:
Step 1: Find the initial vertical velocity
Given:
- Height of the wall (h) = 17.0 m
- Angle of projection (θ) = 35°
We can use the equation of motion to find the initial vertical velocity (v_y0):
h = (v_y0 * t) - (0.5 * g * t^2)
where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time of flight
Since the ball hits the wall at the same height it was hit, the time of flight is the same as the time taken to reach the wall. Let's denote this time as t_w.
Therefore,
17.0 m = (v_y0 * t_w) - (0.5 * g * t_w^2) (equation 1)
Step 2: Find the initial horizontal velocity
Given:
- Distance to the wall (d) = 127.0 m
- Angle of projection (θ) = 35°
- Initial vertical velocity (v_y0) = calculated in step 1
We can use the equation of motion to find the initial horizontal velocity (v_x0):
d = v_x0 * t_w (equation 2)
Step 3: Find the initial speed
Given:
- Initial vertical velocity (v_y0) = calculated in step 1
- Initial horizontal velocity (v_x0) = calculated in step 2
We can find the initial speed (v_0) using the Pythagorean theorem:
v_0 = sqrt(v_x0^2 + v_y0^2)
Now, let's calculate these values step by step:
Step 1: Find the initial vertical velocity (v_y0):
Using equation 1,
17.0 m = (v_y0 * t_w) - (0.5 * g * t_w^2) (equation 1)
Step 2: Find the initial horizontal velocity (v_x0):
Using equation 2,
127.0 m = v_x0 * t_w (equation 2)
Step 3: Find the initial speed (v_0):
Using the Pythagorean theorem,
v_0 = sqrt(v_x0^2 + v_y0^2)
Let's calculate these values one by one:
Step 1: Find the initial vertical velocity (v_y0):
Rearranging equation 1:
(v_y0 * t_w) - (0.5 * g * t_w^2) = 17.0 m
Step 2: Find the initial horizontal velocity (v_x0):
Rearranging equation 2:
v_x0 = 127.0 m / t_w
Step 3: Find the initial speed (v_0):
Using the Pythagorean theorem:
v_0 = sqrt(v_x0^2 + v_y0^2)
Now, let's proceed with the calculations. Please give me a moment.
To solve this problem, we can use the equations of projectile motion.
(a) First, let's find the initial speed of the ball. We know that the ball was hit at an angle of 35° to the horizontal. The initial speed can be found using the horizontal and vertical components of the velocity.
The horizontal component of the velocity (vx) remains constant throughout the motion. We can use the formula:
vx = initial speed * cos(angle)
Substituting the given angle of 35°, we have:
vx = initial speed * cos(35°)
Now, let's find the vertical component of the velocity (vy). It can be found using the formula:
vy = initial speed * sin(angle)
Substituting the given angle of 35°, we have:
vy = initial speed * sin(35°)
Since the ball is hit at a height of 1.0 m, the initial vertical velocity (v0y) is the negative of the final vertical velocity (vy) when the ball reaches the wall. Therefore, we have:
v0y = -vy
Using the values, we can substitute:
v0y = -initial speed * sin(35°)
Now, we can use the relationship between time (t) and vertical displacement (d) to find d when the ball reaches the wall. We know that the vertical displacement is the difference between the height of the wall and the initial height of the ball. Therefore:
d = wall height - initial height
d = 17.0 m - 1.0 m
d = 16.0 m
The vertical displacement can be calculated using the formula:
d = v0y * t + (1/2) * (-9.8 m/s^2) * t^2
Substituting the known values, we have:
16.0 m = (-initial speed * sin(35°)) * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying this equation, we get:
-4.9 t^2 - (initial speed * sin(35°)) * t + 16.0 m = 0
This is a quadratic equation in terms of t. We can solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -4.9, b = -(initial speed * sin(35°)), and c = 16.0.
(b) Once we find the value of t, it will give us the time it takes for the ball to reach the wall.
(c) To find the components of the velocity and the speed of the ball when it reaches the wall, we can use the equations:
vf = sqrt(vx^2 + vy^2)
vx,f = vx
vy,f = vy
Let's plug the values into the equations and solve them step-by-step to find the answers.