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.

Answer 1

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

Answer 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

Answer 3

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