A baseball thrown at an angle of 65.0^\circ above the horizontal strikes a building 16.0 {\rm m} away at a point 5.00 {\rm m} above the point from which it is thrown. Ignore air resistance.
a) find the magnitude of the initial velocity of the baseball.
b) find the magnitude of the velocity of the baseball just before it strikes the building.
c) find the direction of the velocity of the baseball just before it strikes the building.
To solve this problem, we can break it down into three parts:
a) Finding the initial velocity of the baseball.
b) Finding the velocity of the baseball just before it strikes the building.
c) Finding the direction of the velocity of the baseball just before it strikes the building.
To find the initial velocity (magnitude and direction) of the baseball, we can use the following equation:
R = (V^2 * sin(2θ)) / g
Where:
- R is the horizontal range (16.0 m),
- V is the initial velocity of the baseball,
- θ is the angle above the horizontal (65.0°), and
- g is the acceleration due to gravity (9.8 m/s^2).
To find the magnitude of the initial velocity (V), we can rearrange the equation:
V = sqrt((R * g) / sin(2θ))
Substituting the given values, we can solve for V:
V = sqrt((16.0 m * 9.8 m/s^2) / sin(2(65.0°)))
Now, let's calculate V:
V = sqrt((156.8 m^2/s^2) / sin(130.0°))
≈ sqrt(1.2072 m^2/s^2)
≈ 1.098 m/s
Therefore, the magnitude of the initial velocity of the baseball is approximately 1.098 m/s.
To find the velocity of the baseball just before it strikes the building, we can use the equations of motion in the vertical and horizontal directions.
In the vertical direction, the final velocity (Vf) can be found using the equation:
Vf = V0y - g * t
Where:
- V0y is the initial vertical component of the velocity (V0 * sin(θ)),
- g is the acceleration due to gravity (9.8 m/s^2), and
- t is the time taken.
The time (t) can be found using the equation:
t = d / (V0 * cos(θ))
Where:
- d is the vertical displacement (5.00 m),
- V0 is the initial velocity of the baseball, and
- θ is the angle above the horizontal (65.0°).
Substituting the given values, we can solve for t:
t = (5.00 m) / (1.098 m/s * cos(65.0°))
Now, let's calculate t:
t = 5.00 m / (1.098 m/s * 0.4226)
≈ 11.228 s
Next, substituting the values of V0y and t into the equation for Vf, we can solve for Vf:
Vf = (V0 * sin(θ)) - (g * t)
Vf = (1.098 m/s * sin(65.0°)) - (9.8 m/s^2 * 11.228 s)
≈ 6.2854 m/s - 109.0299 m/s
≈ -102.7445 m/s
Therefore, the magnitude of the velocity of the baseball just before it strikes the building is approximately 102.7445 m/s in the downward direction.
To find the direction of the velocity of the baseball just before it strikes the building, we can find the angle (θf) using the equation:
θf = atan(Vfy / Vfx)
Where:
- θf is the final angle of the velocity just before striking the building,
- Vfy is the vertical component of the velocity just before striking the building,
- Vfx is the horizontal component of the velocity just before striking the building.
Since the baseball is only moving horizontally just before it strikes the building, the horizontal component of the velocity (Vfx) remains constant throughout the motion. Thus, Vfx is equal to the initial horizontal component of the velocity (V0 * cos(θ)).
Substituting the known values, we can solve for Vfx:
Vfx = V0 * cos(θ)
= 1.098 m/s * cos(65.0°)
≈ 0.535 m/s
Now, let's calculate θf:
θf = atan(-102.7445 m/s / 0.535 m/s)
≈ -84.415°
Therefore, the direction of the velocity of the baseball just before it strikes the building is approximately 84.415° below the horizontal.