Math Problem Statement

  1. A boy throws a ball horizontally from the top of a building. The initial speed of the ball is 31m/s. Find the horizontal and vertical velocity of the ball after 22s.

Step by step solution solving

Solution

Let's go through the problem step by step to find the horizontal and vertical velocity of the ball after 22 seconds.

Step 1: Understand the problem

  • A ball is thrown horizontally, which means the initial vertical velocity is zero.
  • The only force acting on the ball in the vertical direction is gravity.

Step 2: List the given information

  • Initial horizontal speed: vx0=31m/sv_{x0} = 31 \, \text{m/s}
  • Initial vertical speed: vy0=0m/sv_{y0} = 0 \, \text{m/s}
  • Time after which we need to find the velocities: t=22st = 22 \, \text{s}
  • Acceleration due to gravity: g=9.8m/s2g = 9.8 \, \text{m/s}^2

Step 3: Calculate the horizontal velocity

The horizontal velocity remains constant throughout the motion because there is no acceleration in the horizontal direction (assuming no air resistance).

vx=vx0=31m/sv_x = v_{x0} = 31 \, \text{m/s}

Step 4: Calculate the vertical velocity

The vertical velocity is affected by gravity. The vertical velocity at any time tt can be calculated using the equation:

vy=vy0+g×tv_y = v_{y0} + g \times t

Since the initial vertical velocity vy0v_{y0} is zero, the equation simplifies to:

vy=0+(9.8m/s2)×22sv_y = 0 + (9.8 \, \text{m/s}^2) \times 22 \, \text{s}

vy=215.6m/sv_y = 215.6 \, \text{m/s}

Step 5: Summarize the results

  • Horizontal velocity after 22 seconds: vx=31m/sv_x = 31 \, \text{m/s}
  • Vertical velocity after 22 seconds: vy=215.6m/sv_y = 215.6 \, \text{m/s}

Conclusion

After 22 seconds, the ball has a horizontal velocity of 31 m/s and a vertical velocity of 215.6 m/s.

Would you like to ask anything else? Here are 5 related questions:

  1. How can you find the total velocity of the ball after 22 seconds?
  2. What would be the ball's velocity if it were thrown at an angle instead of horizontally?
  3. How do you determine the height of the building if the time of flight is known?
  4. Can you explain how to calculate the range of the projectile?
  5. How does the time of flight depend on the initial speed and angle of projection?

Tip: Always remember that in the absence of air resistance, the horizontal component of velocity remains constant, while the vertical component changes due to gravity.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics

Formulas

Vertical Velocity: vy = vy0 + gt

Theorems

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Suitable Grade Level

High School