A boy can throw a ball a maximum horizontal distance of R on a level field. How far can he throw the same ball vertically upward? Assume that his muscles give the ball the same speed in each case. (Use R and g as appropriate in your equation.)

R/2

Saferi

To determine the maximum vertical distance, we can use the concept of projectile motion. In projectile motion, the horizontal and vertical components of motion are independent of each other.

Given that the boy can throw the ball a maximum horizontal distance of R, we can use the equation for horizontal displacement:

Horizontal displacement (R) = initial horizontal velocity (Vx) * time of flight (t)

Since there is no acceleration in the horizontal direction (assuming no air resistance), the initial horizontal velocity (Vx) remains constant throughout the flight. Therefore, we can rewrite the equation as:

R = Vx * t

Now let's consider the vertical motion. The ball is thrown upward, against the force of gravity. The only force acting on the ball is gravity, causing it to decelerate until it reaches its maximum height and starts falling down.

Vertical displacement = initial vertical velocity (Vy) * time of flight (t) + (1/2) * acceleration due to gravity (g) * time of flight (t)^2

We want to find the vertical displacement when the ball reaches its maximum height. At this point, the vertical velocity becomes zero. So, we can write:

0 = Vy + g * t

Rearranging the equation, we get:

t = -Vy / g

Substituting this value of time into the vertical displacement equation, we have:

Vertical displacement = Vy * (-Vy / g) + (1/2) * g * (-Vy / g)^2

Simplifying further, we get:

Vertical displacement = (Vy^2) / (2 * g)

Since the boy gives the ball the same speed (magnitude of velocity) in each case, the vertical and horizontal components of velocity (Vy and Vx) are equal, i.e., Vy = Vx.

Therefore, the maximum vertical displacement is:

Vertical displacement = (Vx^2) / (2 * g)

Substituting the given maximum horizontal distance (R), the equation becomes:

Vertical displacement = (R^2) / (2 * g)

Hence, the maximum vertical distance the boy can throw the ball is (R^2) / (2 * g).

To solve this problem, we need to use the principles of projectile motion. In projectile motion, the horizontal and vertical motions are independent of each other.

Let's start by considering the horizontal motion. The boy can throw the ball a maximum horizontal distance of R. This means that the initial horizontal velocity of the ball is referred to as Vx and remains constant throughout the motion.

Now, let's move on to the vertical motion. When the ball is thrown vertically upward, the only force acting on it is the force of gravity, which pulls the ball downward. So, the only acceleration acting on the ball is due to gravity (denoted as g).

The initial vertical velocity of the ball (Vy) is zero because the ball is thrown vertically upward. As the ball moves upward, its vertical velocity decreases until it reaches its maximum height, at which point its vertical velocity becomes zero again. At this point, the ball starts to fall downward due to the force of gravity.

Using the principles of kinematics, we can determine the maximum height reached by the ball. We know that the initial vertical velocity (Vy) is zero, the acceleration (a) is -g (since it acts in the opposite direction to the positive vertical axis), and the final vertical velocity (Vf) is also zero at the maximum height.

The equation that relates these variables is given by:
Vf^2 = Vy^2 + 2aΔy

Since Vy is zero at both the initial and final positions, the equation simplifies to:
0 = 0 + 2*(-g)*Δy
0 = -2gΔy

Simplifying further, we find:
Δy = 0

This means that the maximum height reached by the ball is zero, which implies that the ball does not go vertically upward at all.

Therefore, we can conclude that the boy can throw the same ball a maximum horizontal distance of R on a level field, but he cannot throw it vertically upward to reach any significant height.