A ball is thrown straight up from the edge of the roof of a building. A second b

Question

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03s later. You may ignore air resistance.

a) If the height of the building is 19.1s , what must the initial speed be of the first ball if both are to hit the ground at the same time?

i got 8.36 m/s which is correct

b) Consider the same situation, but now let the initial speed v_0 of the first ball be given and treat the height h of the building as an unknown. What must the height of the building be for both balls to reach the ground at the same time for v_o = 9.00s m/s .

c) If is greater than some value v_max, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for v_max.

d)If v_0 is less than some value v_min, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for v_min.

Explanation / Answer

assumig vertically upward is the positive direction

let ball one is dropped at 0 sec
let both ball hit the ground at t= t sec

time of flight of ball 1 = t sec
time of flight of ball 2 = (t-1.03) sec

h= u1*t - 0.5gt2 = 9*t-0.5*9.8(t)2 1 refers ball 1,

h = u2 *(t-1.03) - 0.5g(t-1.03)2 = 0 -0.5*9.8(t-1.03)2   2 refers to the ball 2

equating

we get

9*t-0.5*9.8(t)2= 0 -0.5*9.8(t-1.03)2

9t= 10.09 t - 5.19

t = 5.19/1.09

solving for t we get t = 4.75 s

h= -67.8 m (- sign means that h is opposite to the direction of initial velocity of ball 2

C)

assuming u as the initial velocity of ball 2

putting in the equations and equating h

we get u*t - 4.9 t2= -4.9 (t-1.03)2

u*t= -5.18 + 10.09*t

(10.094-)t= 5.19

t= (5.18)/(10.094-u) now t >1.03

case 1 u <10.094 otherwise t < 0 not possible so u max = 10.094 m/s

case 2

(5.18)/(10.094-u)>1.03

10.094-u<5.19/1.03

10.094-u < 5.02

u>4.90

so u min = 4.90

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