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.
the range equation is ... R = [v^2 * sin(2Θ)] / g
... v is the launch velocity ... Θ is the launch angle
the sine term is maximum with a 45º launch angle ... sin(90º) = 1
R = v^2 / g ... v = √(R * g)
m * g * h = 1/2 * m * v^2 = 1/2 * m * R * g
h = 1/2 * R
R_{max}= v_{o}^{2}sin2theta/2g
We have the maximum R if sin2theta =1
then our theta will be 45^{0}
so,
Now, v_{yf}^{2} = v_{y0}^{2} - 2g(y_{f} - y_{i})
y = [v_{o}^{2} - v_{yf}^{2} ) /2g]
v_{yf}^{2} = 0
y = [v_{o}^{2} /2g]
We have R= v_{o}^{2}/29
y = R /2
well, let's assume he throws it from the ground level.
If ignoring friction, maximum range is when angle from horizontal to start is 45 degrees (that is a separate problem)
assume speed s
Vi = s sin 45 = .707 s
initial Ke up = (1/2 )m (.707s)^2 = .25 m s^2
potential energy at height h = m g h
so
max height =.25 s^2/g
now how long did it take t get there?
average speed up = .707 s/2 = .3535 s
so time = .25 s^2/g / .3535s = .707 s/g
total time in air = twice that = 2 * .707 s/g
distance = s cos 45 * t = .707 s * 2 * .707 s/g = s^2/g = R
now straight up
(1/2) m s^2 = m g h
h = .5 s^2/g = .5 R
Well, the distance the ball goes vertically upward depends on the gravitational force working against it. So, if we ignore air resistance and other factors, the maximum vertical distance the boy can throw the ball is directly proportional to the maximum horizontal distance he can throw.
But hey, let's give it another twist! If the boy throws the ball vertically upward, it will eventually make a beautiful arc and come crashing back down, thanks to gravity. So, if you really want to see the ball go up and up, I suggest finding a nice, sunny day, attaching the ball to a hot air balloon, and watching it soar into the sky. Just make sure you have a good grip on the string, or else you'll be floating away with it!
To determine the maximum vertical distance the boy can throw the ball upward, we need to understand the relationship between the horizontal and vertical motion of the ball.
When the ball is thrown horizontally, there is no upward velocity component initially, and the only force acting on it is gravity. However, when the ball is thrown vertically upward, both the initial upward velocity and the gravitational force affect its motion.
In both cases, the speed at which the ball is thrown is the same. The horizontal speed does not affect the upward motion of the ball, so it remains constant.
When the ball is thrown upward, it reaches its highest point where its vertical velocity becomes zero. At this point, gravity slows the ball down until it reaches its maximum height and starts falling back down.
To determine the maximum vertical distance, we need to calculate the time it takes for the ball to reach the highest point of its trajectory. This can be done using the equation:
t = v / g
where "t" is the time in seconds, "v" is the initial vertical velocity, and "g" is the acceleration due to gravity (approximately 9.8 m/s² on Earth).
Since the boy throws the ball with the same speed in both cases, we can assume the initial vertical velocity in the upward motion is the same as the initial horizontal velocity in the horizontal motion.
Now, to determine the time it takes for the ball to reach its highest point, we need to find the vertical component of the initial velocity. This can be done using trigonometry.
Let's assume the angle at which the boy throws the ball with respect to the horizontal is θ. Then, the vertical component of the initial velocity (v\_y) can be calculated using:
v\_y = v * sin(θ)
Now we can calculate the time it takes for the ball to reach its highest point:
t = v\_y / g = (v * sin(θ)) / g
Once the ball reaches its highest point, it starts falling back down. The total time for the ball to reach the ground again would be equal to twice the time it took to reach the maximum height:
total time = 2 * t = 2 * (v * sin(θ)) / g
Now, we can calculate the maximum vertical distance (H) by multiplying the total time by the initial vertical velocity:
H = v\_y * total time = 2 * (v² * sin(θ) * cos(θ)) / g = 2 * (v² * sin(2θ)) / g
So, the maximum vertical distance the boy can throw the ball upward is given by:
H = 2 * (v² * sin(2θ)) / g
Note that the actual value of H will depend on the angle at which the boy throws the ball, θ.