Consider a boy standing on a building that is 500m height and another boy standing on the ground. If the boy on the ground throws a ball to the boy standing on the building, with what vertical velocity must the ball be thrown such that it reaches the boy standing on the building. Neglect the height of the boys.
The kinetic energy of the ball when thrown must equal the potential energy it must gain to reach the top.
(1/2) M V^2 = M g H
V = sqrt(2 g H) = 99 m/s = 221 mph
That building is higher than the Empire State Building.
No one can throw a ball that fast. The best fast ball pitchers throw baseballs at 100 mph
Well, let's take a moment to appreciate this situation. One boy on the ground, another boy on a building. The ground is jealous, I tell you. Anyway, back to the question. In order to calculate the required vertical velocity for the ball to reach the boy on the building, we can use the lovely laws of physics.
Considering the height of the building is 500m, we can use the equation of motion:
Δy = v₀t + (1/2)at²
Where:
Δy is the change in height (500m),
v₀ is the initial vertical velocity (what we're trying to find),
t is the time it takes to reach the top, and
a is the acceleration due to gravity (-9.8 m/s²).
Since the ball will start from the ground (with zero initial vertical velocity), we can simplify the equation to:
Δy = (1/2)at²
Plugging in the values, we get:
500 = (1/2)(-9.8)t²
Now, let's solve for t. Dividing both sides by -4.9 (the (1/2)(-9.8)), we get:
t² = -500 / -4.9
t² = 102.04
t ≈ 10.1 seconds
Now that we have the time it takes for the ball to reach the top, we can find the required vertical velocity:
v₀ = Δy / t
v₀ = 500 / 10.1
v₀ ≈ 49.5 m/s
So, the ball must be thrown vertically with an initial velocity of approximately 49.5 m/s to reach the top of the building. Don't forget to aim carefully, though. We don't want any accidental window smashing!
To calculate the vertical velocity required for the ball to reach the boy standing on the building, we can utilize the equations of motion.
We will assume that there is no air resistance.
Given:
Height of the building, h = 500m
Since the initial vertical velocity and final vertical velocity at the building are zero (the ball starts and stops at rest vertically), we can use the equation of motion for vertical displacement:
h = (1/2) * g * t^2,
where:
h = vertical displacement (500m in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time of flight
Rearranging the equation, we get:
t = sqrt(2h / g)
Substituting the values, we have:
t = sqrt((2 * 500) / 9.8)
t ≈ 10.2041 seconds
Now, to find the initial vertical velocity (u), we can use the equation of motion:
v = u + gt,
where:
v = final vertical velocity (0 m/s)
u = initial vertical velocity (what we want to find)
Rearranging the equation, we get:
u = v - gt
u = 0 - (9.8 * 10.2041)
u ≈ -100 m/s
Therefore, the ball must be thrown vertically upwards with an initial vertical velocity of approximately 100 m/s.
To determine the required vertical velocity of the ball thrown by the boy on the ground, we can apply the principles of projectile motion. We need to find the initial vertical velocity component, as the horizontal velocity does not affect the vertical motion.
In this scenario, the initial vertical position of the ball is 0 m (since the ball is being thrown from the ground) and the final vertical position is 500 m (where the boy on the building is located). Since both positions are vertically aligned, we can take the displacement as 500 m.
Also, we know that the acceleration due to gravity (g) is -9.8 m/s² (negative value because it acts in the opposite direction of the positive y-axis).
Using the kinematic equation:
Δy = V₀y * t + (1/2) * a * t²
Where:
Δy = displacement (500 m)
V₀y = initial vertical velocity
a = acceleration due to gravity (-9.8 m/s²)
t = time of flight
Plugging in the values, the equation becomes:
500 m = V₀y * t - 4.9 m/s² * t²
Simplifying the equation:
4.9 t² - V₀y t + 500 = 0
Now, we have a quadratic equation in terms of time. Since we want to find the initial vertical velocity (V₀y), we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
Where:
a = 4.9
b = -V₀y
c = 500
Therefore,
t = (-(-V₀y) ± √((-V₀y)² - 4 * 4.9 * 500)) / (2 * 4.9)
t = (V₀y ± √(V₀y² - 9800)) / 9.8
To ensure that the ball reaches the boy on the building, the time of flight (t) must be positive. Thus, we can ignore the negative solution.
Now, for the ball to reach its highest point, the time of flight should be half of the total time. Therefore:
t_total = 2t
Substituting the value of t from the quadratic formula:
t_total = 2((V₀y + √(V₀y² - 9800)) / 9.8)
Simplifying the expression further:
t_total = (2V₀y + 2√(V₀y² - 9800)) / 9.8
To reach the highest point, the ball needs to have a symmetric time of flight. Thus:
t_total/2 = (V₀y + √(V₀y² - 9800)) / 9.8
Now, to find the required vertical velocity of the ball (V₀y) so that it reaches the boy on the building, we can substitute the known values:
(V₀y + √(V₀y² - 9800)) / 9.8 = (500 / 2)
(V₀y + √(V₀y² - 9800)) = 2450
Squaring both sides of the equation:
V₀y² + 2V₀y√(V₀y² - 9800) + (V₀y² - 9800) = 6012500
2V₀y√(V₀y² - 9800) = 6012500 - 2V₀y² + 9800
Simplifying further:
2V₀y√(V₀y² - 9800) = -2V₀y² + 6022300
Squaring both sides again:
4V₀y⁴ - 47799200V₀y² + 35721090000 = 0
This fourth-degree equation can be solved numerically to find the value(s) of V₀y. However, as it involves complex calculations and cumbersome steps, we can use computational tools or solver software to find a numerical solution.
Hence, the vertical velocity (V₀y) of the ball thrown by the boy on the ground to reach the boy on the building can be determined by solving the quadratic equation or using numerical methods.