A major leaguer hits a baseball so that it leaves the bat at a speed of 33.0 m/s and at an angle of 38.5° above the horizontal. Ignore air resistance.

(a) At what two times is the baseball at a height of 10.7 m above the point at which it left the bat?
(b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a).
(c) What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?

(a) 33 sin38.5° t - 4.9t^2 = 10.7

(b) vertical: 33 sin38.5° - 9.8t
horizontal: 33 cos38.5°
(c) same speed, but negative angle

(a) Well, to find the two times when the baseball is at a height of 10.7 m above the point it left the bat, we'll have to do some math. But don't worry, I'll try to make it as entertaining as possible. So, let's get calculating!

First, we need to figure out the time it takes for the baseball to reach its maximum height (when its vertical velocity is zero). We can use the following equation:

vf = vi + at

Since the baseball's final vertical velocity (vf) is zero, and its initial vertical velocity (vi) is given by the formula vi = v * sin(theta), where v is the speed and theta is the angle, we can rewrite the equation as:

0 = v * sin(theta) + gt

Here, g is the acceleration due to gravity. But since we're ignoring air resistance (which can be a bit rebellious), we can assume g is approximately 9.8 m/s^2, which is the acceleration due to gravity on Earth.

Solving for t, we get:

t = -v * sin(theta) / g

Now, let's plug in the values:

t = -(33.0 m/s) * sin(38.5°) / (9.8 m/s^2)

Calculate that and you'll find the time it takes for the baseball to reach its maximum height.

Next, to find the time at which the baseball is at a height of 10.7 m above the point it left the bat, we need to consider its vertical motion. We can use the following equation:

y = yi + vi * t + (1/2) * a * t^2

Here, y is the height, yi is the initial height (which is 0 m because we consider the point where the ball left the bat as the reference point), vi is the initial vertical velocity (which is v * sin(theta)), t is the time, and a is the acceleration (which is -g due to gravity).

Setting y = 10.7 m, yi = 0 m, vi = v * sin(theta), and a = -g, we get:

10.7 m = (v * sin(theta)) * t + (1/2) * (-g) * t^2

This is a quadratic equation, my friend! Solve it using the formula for finding the roots of a quadratic equation and voila, you'll have your two times when the ball is at a height of 10.7 m above the point it left the bat.

(b) Now, let's move on to calculating the horizontal and vertical components of the baseball's velocity at each of the two times.

To find the horizontal component of velocity (Vx), we can use the formula:

Vx = v * cos(theta)

Where v is the speed and theta is the angle.

To find the vertical component of velocity (Vy), we can use the formula:

Vy = vi + gt

Where vi is the initial vertical velocity and g is the acceleration due to gravity.

Plug in the values for v, theta, vi, and g, and you'll get the horizontal and vertical components of the velocity at each time.

(c) Lastly, the magnitude of velocity is given by calculating the total velocity, which is the hypotenuse of a right triangle formed by the horizontal and vertical components of the velocity. You can use the Pythagorean theorem to find it:

velocity = sqrt(Vx^2 + Vy^2)

Additionally, because you're interested in the direction, you can use the arctan of Vy/Vx, taking into account the appropriate quadrant.

Now, go forth and calculate away! If you need any further clarification or a dose of humor, I'm here to assist you. Good luck, mathlete!

To solve this problem, we can use the equations of motion for projectile motion. Let's break it down step by step:

Step 1: Find the time it takes for the baseball to reach a height of 10.7 m.

We know that the initial vertical velocity is given by V * sin(theta), where V is the initial velocity (33.0 m/s) and theta is the angle (38.5°). The vertical displacement is 10.7 m.

Using the equation:

h = Vt * sin(theta) - (1/2) * g * t^2

Where h is the height, V is the initial velocity, t is the time, theta is the angle, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values:

10.7 = 33.0 * t * sin(38.5°) - (1/2) * 9.8 * t^2

Simplifying:

4.9 t^2 - 17.7 t + 10.7 = 0

This is a quadratic equation that we can solve for t. We can either use the quadratic formula or factor it. Let's factor it:

(2.45t - 7)(2.45t - 1.5) = 0

From this, we can see that t = 7/2.45 = 2.86 seconds or t = 1.5/2.45 = 0.61 seconds.

So, the baseball will reach a height of 10.7 m at two times: 2.86 seconds and 0.61 seconds.

Step 2: Calculate the horizontal and vertical components of the baseball's velocity at each of the two times.

The horizontal velocity remains constant throughout the motion, so it is simply V * cos(theta). Let's calculate it:

Horizontal velocity = 33.0 m/s * cos(38.5°)
= 26.6 m/s

The vertical component of the velocity changes due to the effect of gravity. We can use the equation:

Vf = Vi + gt

Where Vf is the final velocity, Vi is the initial velocity, g is the acceleration due to gravity, and t is the time.

Let's calculate the vertical component of the velocity at both times:

At t = 2.86 seconds:
Vertical velocity = V * sin(theta) + g * t
= 33.0 m/s * sin(38.5°) + 9.8 m/s^2 * 2.86 s
= 20.1 m/s

At t = 0.61 seconds:
Vertical velocity = V * sin(theta) + g * t
= 33.0 m/s * sin(38.5°) + 9.8 m/s^2 * 0.61 s
= 14.0 m/s

So, at t = 2.86 seconds, the horizontal component of the velocity is 26.6 m/s and the vertical component is 20.1 m/s.
At t = 0.61 seconds, the horizontal component of the velocity is 26.6 m/s and the vertical component is 14.0 m/s.

Step 3: Find the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat.

When the baseball returns to the level at which it left the bat, its vertical velocity will be the same as its initial vertical velocity. The horizontal velocity remains constant throughout the motion.

So, the magnitude of the velocity at this point is given by:

Magnitude of velocity = sqrt[(horizontal velocity)^2 + (vertical velocity)^2]
= sqrt[(26.6 m/s)^2 + (20.1 m/s)^2]
= sqrt[707.56 + 404.01]
= sqrt(1111.57)
= 33.37 m/s (rounded off to two decimal places)

The direction of the velocity can be found using the inverse tangent of the vertical component divided by the horizontal component:

Direction = arctan(vertical velocity / horizontal velocity)
= arctan(20.1 m/s / 26.6 m/s)
= arctan(0.756)
= 38.0 degrees (rounded off to one decimal place)

Therefore, when the baseball returns to the level at which it left the bat, its velocity magnitude is 33.37 m/s and its direction is 38.0 degrees.

To solve this problem, we can use projectile motion equations to analyze the motion of the baseball.

(a) At what two times is the baseball at a height of 10.7 m above the point at which it left the bat?

To find the times when the baseball is at a height of 10.7 m, we use the projectile motion equation for vertical displacement:

Δy = v₀y * t + (1/2) * g * t²

Where:
- Δy is the vertical displacement (10.7 m)
- v₀y is the initial vertical component of velocity (v₀ * sin(θ))
- t is the time
- g is the acceleration due to gravity (-9.8 m/s²)

Rearranging the equation, we get:

0 = (1/2) * g * t² + (v₀y * t - Δy)

We can solve this quadratic equation to find the two values of t when the baseball is at a height of 10.7 m.

(b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a).

To calculate the horizontal and vertical components of velocity, we use the following equations:

v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)

Where:
- v₀x is the initial horizontal component of velocity
- v₀y is the initial vertical component of velocity
- v₀ is the initial velocity (33.0 m/s)
- θ is the angle above the horizontal (38.5°)

(c) What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?

To find the velocity when the baseball returns to the level it left the bat, we can use the same magnitude and direction of the initial velocity.

Now, let's calculate the answers step by step.

(a) To find the times when the baseball is at a height of 10.7 m, we solve the quadratic equation:

(1/2) * g * t² + (v₀y * t - Δy) = 0

Plugging in the values:
g = -9.8 m/s²
Δy = 10.7 m
v₀y = v₀ * sin(θ) = 33.0 m/s * sin(38.5°)

We get the quadratic equation:

(1/2) * (-9.8) * t² + (33.0 * sin(38.5°)) * t - 10.7 = 0

Solving this quadratic equation will give us two values of t when the baseball is at a height of 10.7 m.

(b) Once we have the two values of t, we can calculate the horizontal and vertical components of the velocity at each time.

Horizontal component:
v₀x = v₀ * cos(θ)
Where v₀ is the initial velocity (33.0 m/s) and θ is the angle above the horizontal (38.5°).

Vertical component:
v₀y = v₀ * sin(θ)
Where v₀ is the initial velocity (33.0 m/s) and θ is the angle above the horizontal (38.5°).

(c) When the baseball returns to the level it left the bat, the magnitude and direction of its velocity will be the same as the initial velocity. The magnitude of the velocity is 33.0 m/s, and the direction is 38.5° above the horizontal.