A ball player hits a home run, and the baseball just clears a wall 17.0 m high located 124.0 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
a. Dh=Vo^2*sin(2A) / g = 124m = Hor. dist.
Vo^2^sin(79) / 9.8 = 124,
0.9397Vo^2 / 9.8 = 124,
0.9397Vo^2 = 1215.2,
Vo^2 = 1293.2,
Vo = 36m/s @ 35 = Initial velocity.
Xo = 36cos35 = 29.49m/s.
Yo = ver. = 36sin35 = 20.65m/s
c. hmax = (Yf^2 - Yo^2) = 2g,
hmax = (0 - (20.65)^2) / -19.6 = 21.8m.
Yf^2 = Yo^2 + 2g*d,
Yf^2= 0 + 19.6*(21.8-17)=94.1,
Yf = 9.7m/s. = Vy,f.
Xo = 29.49m/s = Vx,f.
Vf = sqrt((29.49)^2 + (9.7)^2) = 31m/s.
b. t(up) = (Yf - Yo) / g,
t(up) = (0 - 20.65) / -9.8 = 2.1071s.
t(dn) = = (Yf - Yo) / g,
t(dn) = (9.7 0) / 9.8 = 0.9898s.
T = t(up) + t(dn)
T = 2.1071 + 0.9898 = 3.1s to reach te
wall.
To solve this problem, we can use the equations of motion in projectile motion.
(a) The initial velocity of the ball can be found by resolving it into horizontal and vertical components. The vertical component of the initial velocity can be found using the initial height:
y = y0 + v0y*t + (1/2)*g*t^2
Where:
y = final vertical position (17.0 m)
y0 = initial vertical position (1.0 m)
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s^2)
Since the ball reaches its maximum height at the halfway point, the time taken to reach the maximum height is equal to the time taken to reach the wall (t/2).
Using the equation, we get:
17.0 = 1.0 + v0y*(t/2) + (1/2)*(-9.8)*(t/2)^2
Simplifying the equation, we get:
16 = v0y*(t/2) - 4.9*(t/2)^2
Using basic algebra, we can find the value of t/2:
v0y = 9.8*(t/2)
Substituting this value back into the equation, we get:
16 = 4.9*(t^2)/4
which simplifies to:
64 = 4.9*t^2
Solving for t, we get:
t = sqrt(64/4.9)
Now, to find v0y:
v0y = 9.8*(t/2) = 9.8*(sqrt(64/4.9)/2)
Calculate this value to find the initial vertical component of the velocity.
For the horizontal component of the initial velocity, we can use the equation:
x = x0 + v0x*t
Where:
x = distance traveled horizontally (124.0 m)
x0 = initial horizontal position (0)
v0x = initial horizontal velocity
Rearranging the equation, we get:
v0x = x/t
Substituting the given values, we can calculate v0x.
Finally, the initial speed can be calculated as the magnitude of the initial velocity, which is given by:
v0 = sqrt(v0x^2 + v0y^2)
Using the calculated values of v0x and v0y, you can find the initial speed.
(b) The time taken for the ball to reach the wall is equal to the time taken for a projectile launched horizontally to travel a horizontal distance equal to the distance to the wall. We can use the equation:
x = x0 + v0*t
Where:
x = distance traveled horizontally (124.0 m)
x0 = initial horizontal position (0)
v0 = initial speed
t = time taken to reach the wall
Rearranging the equation, we get:
t = x/v0
Substituting the given values, we can calculate the time taken.
(c) At the point where the ball reaches the wall, the horizontal component of its velocity will remain constant, and the vertical component will be equal to the initial vertical velocity. Therefore, the horizontal and vertical components of the velocity when the ball reaches the wall are given by:
vy,f = v0y
vx,f = v0x
The speed of the ball when it reaches the wall, which is the magnitude of the velocity vector, can be calculated using the equation:
vf = sqrt(vx,f^2 + vy,f^2)
Calculate these values using the previously derived values of v0x and v0y to find the final components of the velocity and the speed when the ball reaches the wall.
To solve this problem, we can break it down into several steps.
Step 1: Calculate the initial vertical velocity component (Vy).
Given:
- Height (h) = 17.0 m
- Launch angle (θ) = 35°
- Initial height (y0) = 1.0 m
We can use the following equation to determine Vy:
Vy = V0 * sin(θ)
Substituting the given values:
17.0 m - 1.0 m = V0 * sin(35°)
16.0 m = V0 * sin(35°)
V0 = 16.0 m / sin(35°)
V0 ≈ 29.7 m/s
Therefore, the initial vertical velocity component is approximately 29.7 m/s.
Step 2: Calculate the initial horizontal velocity component (Vx).
Given:
- Launch angle (θ) = 35°
- Initial speed (V0) = 29.7 m/s
We can use the following equation to determine Vx:
Vx = V0 * cos(θ)
Substituting the given values:
Vx = 29.7 m/s * cos(35°)
Vx ≈ 24.4 m/s
Therefore, the initial horizontal velocity component is approximately 24.4 m/s.
Step 3: Calculate the time taken to reach the wall (t).
Given:
- Distance (d) = 124.0 m
- Initial horizontal velocity component (Vx) = 24.4 m/s
We can use the following equation to determine t:
d = Vx * t
Rearranging the equation:
t = d / Vx
Substituting the given values:
t = 124.0 m / 24.4 m/s
t ≈ 5.08 s
Therefore, it takes approximately 5.08 seconds for the ball to reach the wall.
Step 4: Calculate the final vertical velocity component (Vy,f).
Given:
- Initial vertical velocity component (Vy) = 29.7 m/s
Since air resistance is negligible, the vertical velocity component does not change. Therefore:
Vy,f = Vy = 29.7 m/s
Therefore, the final vertical velocity component when the ball reaches the wall is 29.7 m/s.
Step 5: Calculate the final horizontal velocity component (Vx,f).
Given:
- Initial horizontal velocity component (Vx) = 24.4 m/s
Since air resistance is negligible, the horizontal velocity component remains constant throughout the motion. Therefore:
Vx,f = Vx = 24.4 m/s
Therefore, the final horizontal velocity component when the ball reaches the wall is 24.4 m/s.
Step 6: Calculate the final speed (Vf).
Given:
- Final vertical velocity component (Vy,f) = 29.7 m/s
- Final horizontal velocity component (Vx,f) = 24.4 m/s
We can use the Pythagorean theorem to determine the final speed:
Vf = sqrt(Vx,f^2 + Vy,f^2)
Substituting the given values:
Vf = sqrt((24.4 m/s)^2 + (29.7 m/s)^2)
Vf ≈ 38.0 m/s
Therefore, the final speed when the ball reaches the wall is approximately 38.0 m/s.