A 0.145 kg baseball is thrown at a launch angle of 30° and strikes the ground at 18 m/s. How fast would it be moving when it reaches the ground if it's launch angle were 45°? Ignore air resistance.
A. 340 m/s
B. 210 m/s
C. 16 m/s
D. 18 m/s
Well, well, well! We have a physics question here! Let's see if I can juggle some numbers for you.
To solve this problem, we can break it down into two components: the vertical and horizontal motions of the baseball.
First, let's tackle the vertical motion. We know that the baseball is launched at 30° and lands on the ground at 18 m/s. So, the velocity in the vertical direction is 18 m/s * sin(30°) – let's call this V_vertical.
Now, let's move on to the horizontal motion. We know that the launch angle is 45°, which means that the horizontal velocity remains unchanged throughout the motion. So, the horizontal velocity is simply 18 m/s * cos(45°) – let's call this V_horizontal.
Now, to find the total velocity when the baseball reaches the ground, we can combine these two components using the Pythagorean theorem. The total velocity (V_total) would be the square root of V_vertical squared plus V_horizontal squared.
So, let me do some mathematical clownery here:
V_total = √(V_vertical^2 + V_horizontal^2)
V_total = √((18 m/s * sin(30°))^2 + (18 m/s * cos(45°))^2)
Now, let me do the calculations... Drumroll, please...
After juggling the numbers, I found that the correct answer is B. 210 m/s.
So, that's the final answer, my friend! So, if you ever find yourself launching baseballs at wild angles, make sure to impress your friends with this knowledge.
To solve this problem, we can use the concept of projectile motion. The initial velocity of the baseball can be broken down into horizontal and vertical components.
Given:
Mass of the baseball (m) = 0.145 kg
Launch angle (θ1) = 30°
Velocity when the baseball strikes the ground (v1) = 18 m/s
Step 1: Find the initial vertical velocity (v1y) using the launch angle and the initial velocity.
v1y = v1 * sin(θ1)
v1y = 18 m/s * sin(30°)
v1y = 18 m/s * 0.5
v1y = 9 m/s
Step 2: Find the initial horizontal velocity (v1x) using the launch angle and the initial velocity.
v1x = v1 * cos(θ1)
v1x = 18 m/s * cos(30°)
v1x = 18 m/s * sqrt(3)/2
v1x = 9 * sqrt(3) m/s
Step 3: Find the time of flight (t) using the vertical motion of the baseball.
Using the equation: v1y = g * t
9 m/s = 9.8 m/s^2 * t
t = 9 m/s / 9.8 m/s^2
t ≈ 0.92 s
Step 4: Find the final vertical velocity (v2y) when the baseball reaches the ground.
Using the equation: v2y = v1y - g * t
v2y = 9 m/s - 9.8 m/s^2 * 0.92 s
v2y ≈ 9 m/s - 8.98 m/s
v2y ≈ 0.02 m/s
Step 5: Find the final horizontal velocity (v2x) when the baseball reaches the ground.
Using the equation: v2x = v1x
v2x = 9 * sqrt(3) m/s
Step 6: Find the magnitude of the final velocity (v2) using the final vertical and horizontal components.
Using the Pythagorean theorem: v2^2 = v2x^2 + v2y^2
v2^2 = (9 * sqrt(3))^2 + (0.02)^2
v2^2 = 243 + 0.0004
v2^2 ≈ 243
v2 ≈ sqrt(243)
v2 ≈ 15.59 m/s
Step 7: Comparing the two scenarios, we find that the magnitude of the final velocity when the launch angle is 45° is approximately 15.59 m/s. Therefore, the correct option is:
C. 16 m/s
To solve this problem, we need to use the principles of projectile motion. We'll analyze the horizontal and vertical components of the motion separately.
Let's first consider the vertical motion. We know that the baseball is thrown at a launch angle of 30° and strikes the ground with a final velocity of 18 m/s. Since there is no air resistance, the vertical component of the velocity (V_y) at the peak of the projectile's trajectory will be equal to zero. We can use the kinematic equation for vertical motion to find the initial vertical velocity (V_y0) of the baseball:
Vy = Vy0 + gt,
where Vy is the final vertical velocity (zero in this case), Vy0 is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
In the case of a projectile launched at 30°, the time of flight can be calculated using the equation:
t = 2 * (Vy0 / g).
Since Vy0 = 0 at the peak of the trajectory, the time of flight is given by:
t = 2 * (0 / 9.8) = 0 seconds.
This means that the total time of flight is zero. The baseball will hit the ground immediately after being launched vertically.
Now let's consider the horizontal motion. The horizontal component of the velocity (V_x) remains constant throughout the motion. To find the initial horizontal velocity (V_x0), we can use the equation:
Vx = Vx0,
where Vx is the final horizontal velocity (which is equal to the initial horizontal velocity, V_x0, since there is no horizontal acceleration).
The initial horizontal velocity can be found using the equation:
V_x0 = V0 * cos(θ),
where V0 is the initial velocity of the baseball and θ is the launch angle. Given that the baseball strikes the ground with a final velocity of 18 m/s, we can use this information to find V0:
V0 = 18 m/s.
Now, if the launch angle were 45° instead of 30°, we can calculate the initial horizontal velocity:
V_x0 = V0 * cos(θ),
V_x0 = 18 * cos(45°),
V_x0 = 12.73 m/s.
Therefore, if the launch angle were 45°, the initial horizontal velocity of the baseball would be 12.73 m/s.
Note that the question asks for the speed, not the velocity, when the baseball reaches the ground. Speed is a scalar quantity that only considers the magnitude, not the direction, of the velocity. Since the initial horizontal velocity remains constant throughout the motion, the final speed when the baseball reaches the ground remains the same as the initial horizontal velocity.
Hence, the correct answer is C. 16 m/s.