Math Problem Statement

5. A boy throws a stone from the top of a building 46.0 m above ground. The stone is thrown at an angle of 33.0° below the horizontal and strikes the ground 55.6 m away from the building, find the following: (a) Time of flight. (b) Initial speed. (c) The magnitude and the direction of the velocity of the stone just before it strikes the ground. 6. A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9° above the horizontal. You can ignore air resistance. (a) At what two times is the baseball at a height of 10 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?

Solution

The image you uploaded contains two projectile motion problems.

Problem 5:

  • Situation: A stone is thrown from the top of a building that is 46.0 m high. It is thrown at an angle of 33.0° below the horizontal and strikes the ground 55.6 m away from the building.
  • Questions:
    1. Time of flight: Calculate how long it takes for the stone to hit the ground.
    2. Initial speed: Determine the stone's initial velocity.
    3. Velocity before impact: Find the magnitude and direction of the stone’s velocity just before it hits the ground.

Problem 6:

  • Situation: A baseball is hit with an initial speed of 30.0 m/s at an angle of 36.9° above the horizontal. Air resistance is ignored.
  • Questions:
    1. Time at two heights: Find the two times at which the baseball is at a height of 10 m above the point where it was hit.
    2. Velocity components: Calculate the horizontal and vertical components of the baseball's velocity at each of these two times.
    3. Return velocity: Determine the magnitude and direction of the baseball’s velocity when it returns to the same height from which it was hit.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics
Trigonometry

Formulas

s = ut + 1/2 at^2
v = u + at
Horizontal distance = u * cos(θ) * t
Vertical distance = u * sin(θ) * t - 1/2 * g * t^2
Magnitude of velocity = √(vx^2 + vy^2)
Angle of velocity = tan^(-1)(vy / vx)

Theorems

Equations of motion under constant acceleration
Trigonometric relationships in kinematics

Suitable Grade Level

Grades 11-12