PHYSICS HW HELP PLEASE!!!!! A rubber ball is shot straight up from the ground wi
ID: 2222695 • Letter: P
Question
PHYSICS HW HELP PLEASE!!!!! A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a.)At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g . b.)What is the maximum value of h for which a collision occurs before the first ball falls back to the ground? Express your answer in terms of the variable and appropriate constants. c.)For what value of h does the collision occur at the instant when the first ball is at its highest point? Express your answer in terms of the variable v0 and appropriate constants.Explanation / Answer
PLEASE RATE ME AND AWARD ME KARMA POINTS IF IT IS HELPFUL FOR YOU (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) ----------------------------------------… (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)
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