Math Problem Statement

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.

Solution

Given that a boy throws a ball horizontally from the top of a building with an initial speed of 31m/s31 \, \text{m/s}, we are asked to find the horizontal and vertical velocity of the ball after 22 seconds. We will assume the motion takes place in the absence of air resistance.

Horizontal Velocity:

The horizontal velocity remains constant in the absence of air resistance since there are no forces acting horizontally (assuming ideal conditions). Therefore, the horizontal velocity remains the same as the initial velocity: vx=31m/sv_x = 31 \, \text{m/s}

Vertical Velocity:

The vertical velocity is affected by gravity. The initial vertical velocity vy0v_{y0} is zero since the ball is thrown horizontally. The vertical velocity after time tt is given by: [ v_y = v_{y0} + g \cd

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Math Problem Analysis

Mathematical Concepts

Kinematics
Projectile Motion
Newton's Laws of Motion

Formulas

Horizontal velocity remains constant: v_x = initial velocity
Vertical velocity formula: v_y = v_{y0} + g \cdot t

Theorems

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Suitable Grade Level

Grades 11-12