If you can throw a ball vertically upward to a height h = 30.1 m, what is the maximum horizontal range over which you can throw the same ball, assuming you throw it at the same initial speed

Question

If you can throw a ball vertically upward to a height h = 30.1 m, what is the maximum horizontal range over which you can throw the same ball, assuming you throw it at the same initial speed

Answer 1

Twice as far.

Here's the reason
The maximum height when you throw vertically is V^2/2g.
(That is easily proven using conservation of energy)
The RANGE X at any angle A is
2 sinAcosA/V^2/g, and the maximum range is obtainec at a A = 45 degree angle, for which
Xmax = V^2/g

Answer 2

To determine the maximum horizontal range for a ball thrown vertically, we need to calculate the time it takes for the ball to reach its maximum height and then double that time.

The initial vertical velocity when throwing a ball upward is the same as the final velocity when the ball reaches its highest point, but in the opposite direction. Let's denote this initial velocity as 'v'.

Using the kinematic equation for vertical motion, we can find the time it takes for the ball to reach its highest point:

vf = vi + at

Since the final velocity is 0 m/s at the highest point (as the ball momentarily stops), we have:

0 = v - gt

Simplifying the equation, we get:

t = v / g

where 'g' is the acceleration due to gravity (approximately 9.8 m/s²).

Now that we know the time it takes for the ball to reach its maximum height, we can calculate the total time it takes for the ball to return to its initial height.

Since the ball goes up and then down, the total time required is twice the time calculated above. Let's denote this total time as 'T':

T = 2t

Now we can determine the horizontal range 'R'. The horizontal range can be found using the formula:

R = v * T

Substituting the expression for 'T' that we derived earlier, we get:

R = v * 2t

Finally, substituting the values given in the problem, we can calculate the maximum horizontal range:

R = v * 2 * (v / g)

Given that the initial height 'h' is 30.1 m, the expression becomes:

R = √(2gh) * 2 * (√(2gh) / g)

R = ((2gh)^(1/2)) * 2 * (((2gh)^(1/2)) / g)

Simplifying further:

R = 4h

Substituting the value of 'h' as 30.1 m:

R = 4 * 30.1

R ≈ 120.4 m

Therefore, the maximum horizontal range over which you can throw the ball is approximately 120.4 meters.