Question

A rubber ball is shot straight up from the ground with speed v_0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. 1.) At what height above the ground do the balls collide in terms of v_0, g, and h? 2.) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground in terms of v_0? 3.) For what value of h does the collision occur at the instant when the first ball is at its highest point in terms of v_0?

Answer 1

what is {\kern 1pt}

Answer 2

(A) let the ball meet after t sec,
=>[The distance travelled by the drop ball] + [distance travelled by the thrown ball] = h [in t sec]
=>[by s = ut + 1/2gt^2] + [by s = ut - 1/gt^2] = h
=>[0 + 1/2gt^2] + [Vo x t - 1/gt^2] = h
=>Vo x t = h
=>t = h/Vo ---------------(i)
Let the height at which they meet is H meter from the ground
Thus By s = ut - 1/2gt^2
=>H = Vo x h/Vo - 1/2 x g x (h/Vo)^2
=>H = h - h/(2gVo^2) ----------------------------------------…

Answer 3

B)The drop ball should fall up to the maximum height gained by the thrown ball to meet,
For thrown ball:-
=>By v = u - gt
=>t = Vo/g
=>By v^2 = u^2 - 2gh
=>H = Vo^2/2g
Let the drop ball is at h(max) thus in t sec its height from ground should be h(max) - H
=>By s = ut + 1/2gt^2
=>h(max) - H = 0 + 1/2 x g x (Vo/g)^2
=>h(max) = H + Vo^2/2g
=>h(max) = Vo^2/2g + Vo^2/2g
=>h(max) = Vo^2/g ----------------------------------(Y)
(C)The same as (Y) as this the condition for (B)

Answer 4

The dropped ball has an initial height, so I think its kinematic equation must be represented differently then the one for the ball shot up into the air. In other words,wouldn't we want to use: \[h_{f}=h_{i}+v_{i}\Delta t-\frac{ 1 }{ 2 } a \Delta t ^{2}\] for the ball dropped from above since it has an initial height? ...and what about the initial velocities aren't they both equal to zero so wouldn't it be:
\[h _{f}=h _{i} - \frac{ 1 }{ 2 }a \Delta t ^{2}\] for the ball dropped from above? With \[h _{f}= -\frac{ 1 }{ 2 }a \Delta t ^{2}\] being the kinematic for the ball shot up into the air from below?

Answer 5

Also note that the acceleration rate due to gravity would have the same negative sign for both balls since they are both under free-fall conditions.

Answer 6

Ok, so no one else is helping with this question so far, and if I follow ksaimouli's suggestions I still only end up with\[h_{\max}=h_{i} -g \Delta t^2\]and the (delta)t^2 isn't allowed in the answer since I need to represent this answer only in terms of \[v_{0},~g,~and~h\]
Given that restriction it seems like I should be using this kinematic
\[v_{f}^2=v_{0}^2+2g\Delta h\]instead of the one ksaimouli was suggesting, but the final velocity is also uknown as well as the height so even if I solved this for h for one of the balls its final velocity is still uknown and there is still no solution.

Answer 7

ksaimouli: Your answer for part A was incorrect, the correct answer should have been:
\[h-\frac{ gh^2 }{ 2v_0^2 }\]