A ball player hits a home run, and the baseball just clears a wall 20.3 m high located 119.0 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?
(b) How much time does it take for the ball to reach the wall?
(c) Find the components of the velocity and the speed of the ball when it reaches the wall
vy,f =
vx,f =
vf =
To solve this problem, we can use kinematic equations. Let's break down each part of the problem and solve it step by step.
(a) What is the initial speed?
The initial speed of the ball can be found by decomposing the initial velocity into its horizontal and vertical components. The vertical component of the initial velocity can be found using the equation:
Vy = V * sin(θ)
where V is the initial velocity and θ is the angle of the ball's trajectory. Given that θ = 35°, we can rearrange the equation to solve for V:
V = Vy / sin(θ)
Substituting the values, we get:
V = 1.0 m/s / sin(35°)
V ≈ 1.83 m/s
Therefore, the initial speed of the ball is approximately 1.83 m/s.
(b) How much time does it take for the ball to reach the wall?
To find the time taken by the ball to reach the wall, we need the horizontal component of the velocity. The horizontal component of the initial velocity can be found using the equation:
Vx = V * cos(θ)
Given that θ = 35° and V = 1.83 m/s, we can substitute the values into the equation:
Vx = 1.83 m/s * cos(35°)
Vx ≈ 1.50 m/s
Now, we can use the horizontal velocity to calculate the time taken by the ball to reach the wall. The equation we'll use is:
t = d / Vx
where d is the horizontal distance to the wall. Given that d = 119.0 m, we can substitute the values:
t = 119.0 m / 1.50 m/s
t ≈ 79.33 s
Therefore, it takes approximately 79.33 seconds for the ball to reach the wall.
(c) Find the components of the velocity and the speed of the ball when it reaches the wall.
To find the final velocity of the ball, we can use the equation:
Vf = √(Vx^2 + Vy^2)
Given that Vx = 1.50 m/s and Vy = 1.0 m/s, we can substitute the values:
Vf = √(1.50 m/s)^2 + (1.0 m/s)^2
Vf ≈ √2.25 m^2/s^2 + 1.0 m^2/s^2
Vf ≈ √3.25 m^2/s^2
Vf ≈ 1.80 m/s
Therefore, the speed of the ball when it reaches the wall is approximately 1.80 m/s.
To find the vertical component of the velocity when it reaches the wall, we can use the equation:
Vy,f = Vy = 1.0 m/s
Therefore, the vertical component of the velocity when the ball reaches the wall is 1.0 m/s.
To find the horizontal component of the velocity when it reaches the wall, we can use the equation:
Vx,f = Vx = 1.50 m/s
Therefore, the horizontal component of the velocity when the ball reaches the wall is 1.50 m/s.
In summary:
(a) The initial speed of the ball is approximately 1.83 m/s.
(b) It takes approximately 79.33 seconds for the ball to reach the wall.
(c) The components of the velocity when the ball reaches the wall are:
- vy,f = 1.0 m/s
- vx,f = 1.50 m/s
- vf = 1.80 m/s