Ceiling height for a game of catch. Two boys "play catch"...

Ceiling height for a game of catch. Two boys "play catch" with a ball in a long hallway. The ceiling height is $H$, and the ball is thrown and caught at shoulder height, which we call $h$ for each boy. If the boys are capable of throwing the ball with velocity $v_{0}$, at what maximum separation can they play? Ans. $R=4 \sqrt{(H-h)\left[v_{0}^{2} / 2 g-(H-h)\right]}$ Show that if $H-h>v_{0}^{2} / 4 g, R=v_{0}^{2} / g .$ Explain the physical significance of the condition $H-h>v_{0}^{2} / 4 g$.

Ceiling height for a game of catch. Two boys "play catch" with a ball in a long hallway. The ceiling height is $H$, and the ball is thrown and caught at shoulder height, which we call $h$ for each boy. If the boys are capable of throwing the ball with velocity $v_{0}$, at what maximum separation can they play? Ans. $R=4 \sqrt{(H-h)\left[v_{0}^{2} / 2 g-(H-h)\right]}$
Show that if $H-h>v_{0}^{2} / 4 g, R=v_{0}^{2} / g .$ Explain the physical significance of the condition $H-h>v_{0}^{2} / 4 g$.

Key Concepts

Projectile Motion Kinematics

This concept involves analyzing the motion of an object thrown into the air where its horizontal and vertical components are treated separately. The horizontal motion occurs at a constant velocity while the vertical motion is uniformly accelerated due to gravity, forming a parabolic trajectory. Applying the kinematic equations allows you to determine the time of flight and the range of the projectile given the initial velocity.

Ceiling Constraint in Projectile Motion

When a projectile is thrown in an environment with a height limitation, like a ceiling, its trajectory may be altered if it comes too close to the overhead boundary. The gap between the thrower’s shoulder height (h) and the ceiling (H) restricts the maximum height the ball can reach. The condition H-h > v0^2/(4g) signifies that the available vertical space is sufficient for the ball to complete its natural parabolic arc without being obstructed by the ceiling, thereby allowing the throwers to achieve the maximum range of v0^2/g.

Transcript

00:01 In this question we have given two blocks both have same mass and it is told that both are connected through a pulley and we have given that m is allowed to move down on inclined surface.

00:14 We have to find the tension on this string so first of all i am drawing the diagram for this question so we have given this is an inclined player and there is a mass m on this inclined play and we have given another mass here that is capital m and both are connected through a pulley so we can say that this mass is m and this mass is also m the angle is theta let the tension in this string is that is t so we can say that and the weight of this block is acting vertically downward that is capital mg and the component of this weight in this direction is given by this is mg sine teta now it is told that this block is moving downward let the acceleration of this block in this direction is a so acceleration of this block in this section is also a now i'm going to force balance for this equation so we can say that for this block we can say that for equilibrium for the motion we can say that net force on the block is equal to mass into acceleration we can say that the motion in downward direction so that force is this is mg sine theta minus t and this is equals to m a this is our equation first and for this block we can say that this is our block here on horizontal surface and its acceleration is in this direction ea and the tension force is t.

02:01 So we can say that on this block the net force is tension force and this is equal to mass into acceleration.

02:07 This is our equation second.

02:09 So we can say that if we put the value of t in equation first, so we can say that this is from equation 1 and equation 2.

02:34 We can say that this is m.

02:41 M .g sine theta minus m a is equal to this is m into a...

Please give Ace some feedback