A person can throw a stone to a maximum distance of 100m.the greatest height to which he can throw the stone is?

the simplified range equation is

... Rmax = v² / g ... 100 = v² / g
... 100 g = v²

when an object is thrown straight up, the initial kinetic energy becomes gravitational potential energy
... ½ m v² = m g h ... v² / (2 g) = h

combining the two formulas
... (100 g) / (2 g) = h
... 50 = h

To find the greatest height a person can throw a stone, we can use the formula for projectile motion. The maximum height can be calculated using the formula:

H = (V^2 * sin^2θ) / (2 * g)

Where:
H = maximum height
V = initial velocity
θ = angle of projection
g = acceleration due to gravity (approximately 9.8 m/s^2)

Since the question only provides the maximum distance of 100m, we need to assume an angle of projection. Let's assume an angle of 45 degrees for simplicity.

Using this angle, the equation becomes:

H = (V^2 * sin^2(45)) / (2 * 9.8)

We can rearrange the equation to solve for V:

V^2 = (H * 2 * 9.8) / sin^2(45)
V^2 = (H * 19.6) / 0.5
V^2 = 39.2H

Now, let's substitute the given maximum distance of 100m into the equation:

100 = 39.2H

Solving for H:

H = 100 / 39.2
H ≈ 2.55 meters

Therefore, the greatest height to which the person can throw the stone is approximately 2.55 meters.

To find the greatest height to which the person can throw the stone, we can use the concept of projectile motion. In projectile motion, the maximum height is reached at the halfway point of the total distance.

Given that a person can throw a stone to a maximum distance of 100m, the halfway point would be 50m (half of 100m).

To find the maximum height reached, we can use the following formula for the vertical motion of a projectile:

h = (v^2 * sin^2(θ))/(2 * g)

Where:
- h is the maximum height
- v is the initial velocity
- θ is the angle of projection
- g is the acceleration due to gravity (approximately 9.8 m/s^2)

Since the angle of projection is not given, we can assume an angle of 45 degrees, which would result in the maximum distance for the given initial velocity.

Substituting the values, we have:

h = (v^2 * sin^2(45))/(2 * g)

To find the initial velocity, we can use the horizontal distance:

d = v * cos(θ) * t

Where:
- d is the horizontal distance
- t is the time of flight

Since the stone is thrown horizontally, the vertical component of the initial velocity is 0. Therefore, we can write:

100 = v * cos(45) * t

Simplifying, we have:

100 = (v * √2)/2 * t
200t = v * √2

From here, we need the value of t to solve for v. Using the equation for vertical motion, we can substitute t in terms of v:

t = d / (v * cos(θ))

Substituting the given values:

t = 100 / (v * cos(45))

Simplifying:

t = 100 / (v * √2/2)

t = 200 / (v * √2)

Substituting this value of t back into the equation:

200 * (200 / (v * √2)) = v * √2

40000 / (√2) = v

v ≈ 282.84 m/s

Now that we have the initial velocity, we can substitute it into the initial equation to find the maximum height:

h = (v^2 * sin^2(45))/(2 * g)

h = (282.84^2 * sin^2(45))/(2 * 9.8)

h ≈ 404.94 m

Therefore, the greatest height to which the person can throw the stone is approximately 404.94 meters.