How Fast Must a Volleyball Be Served to Clear the Net and Land Inbounds?

  • Thread starter Thread starter noeinstein
  • Start date Start date
  • Tags Tags
    Magnitude
AI Thread Summary
To determine the required serve speed in women's volleyball, the net height is 2.24 m and the player serves from a height of 3.05 m, needing to clear a horizontal distance of 8.3 m. The minimum initial velocity for the ball to clear the net is calculated using the formula V(o)=R x sqrt(g/2h), where h is the height difference of 0.81 m. The calculated minimum velocity is approximately 16.54 m/s. Additionally, the maximum velocity must ensure the ball lands within the back line after crossing the net. Accurate calculations are essential for effective serving strategies in volleyball.
noeinstein
Messages
14
Reaction score
0

Homework Statement



For women's volleyball the top of the net is 2.24 m above the floor and the court measures 9.0 m on each side of the net. Using a jump serve, a player strikes the ball at a point that is 3.05 m above the floor and a horizontal distance of 8.3 m from the net.

(a) If the initial velocity of the ball is horizontal, what minimum magnitude must it have if the ball is to clear the net?

(b) What maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net?




Homework Equations



V(o)=R x sqrt(g/2h)

The Attempt at a Solution



My wild attempt
H in relation to the top of the net = 3.05m - 2.24m = 0.81m
V(o)= 8.3 x sqrt(9.8/2 x .81) = 16.54
 
Physics news on Phys.org
noeinstein said:

Homework Statement



For women's volleyball the top of the net is 2.24 m above the floor and the court measures 9.0 m on each side of the net. Using a jump serve, a player strikes the ball at a point that is 3.05 m above the floor and a horizontal distance of 8.3 m from the net.

(a) If the initial velocity of the ball is horizontal, what minimum magnitude must it have if the ball is to clear the net?

(b) What maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net?

Homework Equations



V(o)=R x sqrt(g/2h)

The Attempt at a Solution



My wild attempt
H in relation to the top of the net = 3.05m - 2.24m = 0.81m
V(o)= 8.3 x sqrt(9.8/2 x .81) = 16.54
Mostly correct, but V(o)=R x sqrt(g/(2h))

or v_0\,=\,R\sqrt{\frac{g}{2h}}
 
Thank you much sir for your time!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top