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.)

Question

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.)

Answer 1

Duplicate post; already answered. Look elsewhere.

Answer 2

To find the maximum vertical distance the boy can throw the ball upward, we can use the concept of projectile motion. In projectile motion, the horizontal and vertical motions can be considered independently.

We know that the horizontal distance, R, is the maximum distance the ball can travel in the horizontal direction. This implies that the time of flight, tf, is the same for both the vertical and horizontal motions.

In the vertical motion, the ball moves upward against the force of gravity until it reaches its highest point and then starts to fall back down. At the highest point, the vertical velocity, vfy (final vertical velocity), becomes zero.

Using the kinematic equation for vertical motion:

vfy = viy - g * t

Where:
- vfy is the final vertical velocity (zero in this case)
- viy is the initial vertical velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time of flight

We know that viy (initial vertical velocity) is the same as the initial horizontal velocity since the boy throws the ball with the same speed in both cases.

Therefore, viy = vix

Where:
- viy is the initial vertical velocity
- vix is the initial horizontal velocity (same as the speed given in the question)

For the horizontal motion, we can use the equation:

R = vix * tf

Now, we can use the equation for vfy and the equation for R to find the maximum vertical distance, ymax.

First, rearrange the equation for R to solve for tf:

tf = R / vix

Then, substitute the value of tf into the equation for vfy:

0 = vix - g * (R / vix)

Simplifying the equation gives:

vix^2 = g * R

Finally, solve for ymax:

ymax = viy^2 / (2 * g)

But we know from viy = vix that viy = vix, so we can rewrite the equation as:

ymax = vix^2 / (2 * g)

Substituting the previously obtained equation for vix^2:

ymax = (g * R) / (2 * g)

Cancelling out the g factors:

ymax = R / 2

Therefore, the maximum vertical distance the boy can throw the ball upward is half of the maximum horizontal distance, R.