A Baseball Is Thrown At A Velocity Of 87 Ft/s, At An Angle Of 40° Above The Horizontal. Calculate (a) The Maximum Height Of The Ball, (b) The Time In The Air (c) The Horizontal Distance (range) To The Point Where The Ball Strikes The Ground.

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Introduction

In this article, we will explore the physics of a baseball thrown at a velocity of 87 ft/s, at an angle of 40° above the horizontal. We will calculate the maximum height of the ball, the time it spends in the air, and the horizontal distance (range) to the point where the ball strikes the ground. This problem is a classic example of projectile motion, which is a fundamental concept in physics.

Projectile Motion

Projectile motion is a type of motion where an object is thrown or launched into the air, and its trajectory is influenced by the force of gravity. The motion of a projectile can be broken down into two components: horizontal and vertical. The horizontal component is independent of the vertical component, and the object's velocity in the horizontal direction remains constant throughout its flight.

Calculating Maximum Height

To calculate the maximum height of the ball, we need to use the equation for the vertical component of the projectile's motion:

y = v0y * t - (1/2) * g * t^2

where:

  • y is the height of the ball above the ground
  • v0y is the initial vertical velocity of the ball
  • t is the time in the air
  • g is the acceleration due to gravity (approximately 32 ft/s^2)

We can find the initial vertical velocity (v0y) using the equation:

v0y = v0 * sin(θ)

where:

  • v0 is the initial velocity of the ball (87 ft/s)
  • θ is the angle of projection (40°)

Plugging in the values, we get:

v0y = 87 ft/s * sin(40°) = 58.4 ft/s

Now, we can plug in the values into the equation for the vertical component of the motion:

y = 58.4 ft/s * t - (1/2) * 32 ft/s^2 * t^2

To find the maximum height, we need to find the time when the vertical velocity is zero. We can do this by taking the derivative of the equation with respect to time and setting it equal to zero:

dv/dt = 58.4 ft/s - 64 ft/s * t = 0

Solving for t, we get:

t = 58.4 ft/s / 64 ft/s = 0.91 s

Now, we can plug this value back into the equation for the vertical component of the motion:

y = 58.4 ft/s * 0.91 s - (1/2) * 32 ft/s^2 * (0.91 s)^2 = 53.1 ft - 14.4 ft = 38.7 ft

Therefore, the maximum height of the ball is approximately 38.7 ft.

Calculating Time in the Air

To calculate the time in the air, we need to use the equation for the vertical component of the projectile's motion:

y = v0y * t - (1/2) * g * t^2

We can plug in the values and solve for t:

y = 0 (since the ball hits the ground) v0y = 58.4 ft/s g = 32 ft/s^2

Rearranging the equation, we get:

t^2 = 2 * y / g = 2 * 0 / 32 ft/s^2 = 0

Taking the square root of both sides, we get:

t = √0 = 0 s

However, this is not the correct answer. The ball takes some time to hit the ground, so we need to use a different approach.

We can use the equation for the time of flight:

t = 2 * v0y / g

Plugging in the values, we get:

t = 2 * 58.4 ft/s / 32 ft/s^2 = 3.64 s

Therefore, the time in the air is approximately 3.64 s.

Calculating Horizontal Distance (Range)

To calculate the horizontal distance (range), we need to use the equation for the horizontal component of the projectile's motion:

x = v0x * t

where:

  • x is the horizontal distance
  • v0x is the initial horizontal velocity of the ball
  • t is the time in the air

We can find the initial horizontal velocity (v0x) using the equation:

v0x = v0 * cos(θ)

where:

  • v0 is the initial velocity of the ball (87 ft/s)
  • θ is the angle of projection (40°)

Plugging in the values, we get:

v0x = 87 ft/s * cos(40°) = 69.5 ft/s

Now, we can plug in the values into the equation for the horizontal component of the motion:

x = 69.5 ft/s * 3.64 s = 253.1 ft

Therefore, the horizontal distance (range) is approximately 253.1 ft.

Conclusion

Q: What is the difference between the initial velocity and the initial vertical velocity?

A: The initial velocity is the total velocity of the ball, which is the vector sum of the horizontal and vertical components. The initial vertical velocity is the component of the initial velocity that is perpendicular to the horizontal plane.

Q: How do you calculate the initial vertical velocity?

A: To calculate the initial vertical velocity, you need to multiply the initial velocity by the sine of the angle of projection. In this case, the initial vertical velocity is 58.4 ft/s, which is calculated by multiplying the initial velocity (87 ft/s) by the sine of the angle of projection (40°).

Q: What is the significance of the time when the vertical velocity is zero?

A: The time when the vertical velocity is zero is the time when the ball reaches its maximum height. At this point, the ball is momentarily at rest, and its vertical velocity is zero.

Q: How do you calculate the time of flight?

A: To calculate the time of flight, you need to use the equation for the time of flight, which is given by:

t = 2 * v0y / g

where:

  • t is the time of flight
  • v0y is the initial vertical velocity
  • g is the acceleration due to gravity

Q: What is the difference between the time of flight and the time in the air?

A: The time of flight is the time it takes for the ball to reach its maximum height and return to the ground. The time in the air is the time it takes for the ball to travel from the point of projection to the point of impact.

Q: How do you calculate the horizontal distance (range)?

A: To calculate the horizontal distance (range), you need to use the equation for the horizontal component of the projectile's motion, which is given by:

x = v0x * t

where:

  • x is the horizontal distance
  • v0x is the initial horizontal velocity
  • t is the time in the air

Q: What is the significance of the angle of projection?

A: The angle of projection is the angle between the initial velocity and the horizontal plane. It determines the trajectory of the projectile and affects the maximum height, time of flight, and horizontal distance (range).

Q: How do you calculate the maximum height?

A: To calculate the maximum height, you need to use the equation for the vertical component of the projectile's motion, which is given by:

y = v0y * t - (1/2) * g * t^2

where:

  • y is the maximum height
  • v0y is the initial vertical velocity
  • t is the time when the vertical velocity is zero
  • g is the acceleration due to gravity

Q: What is the significance of the acceleration due to gravity?

A: The acceleration due to gravity is the rate at which the ball's velocity changes due to the force of gravity. It affects the maximum height, time of flight, and horizontal distance (range) of the projectile.

Q: How do you calculate the initial horizontal velocity?

A: To calculate the initial horizontal velocity, you need to multiply the initial velocity by the cosine of the angle of projection. In this case, the initial horizontal velocity is 69.5 ft/s, which is calculated by multiplying the initial velocity (87 ft/s) by the cosine of the angle of projection (40°).

Q: What is the significance of the horizontal distance (range)?

A: The horizontal distance (range) is the distance between the point of projection and the point of impact. It is affected by the initial velocity, angle of projection, and acceleration due to gravity.