2. A ball is thrown straight up from the edge of the roof of a building. A secon

Question

2. A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 2.00 s later a) If the height of the building is 6.00 m, what must be the initial speed of the first ball if both are to hit the ground at the same time? Consider the same situation, but now let the initial speedvo of the first ball be given and treat the height h of the building as an unknown b) What must be the height of the building be for both balls to reach the ground at the same time for each of the following values of vo: i) 13.0 m/s ii) 19.2 m/s c) If vo is greater than some value vmar, there is no value of h for which both balls hit the ground at the same time. Solve for vmar. The value vmaz has a simple physical interpretation. What is it? d) If vo is less than some value Umin, there is no value of h for which both balls hit the ground at the same time. Solve for vmin. The value vmin also has a simple physical interpretation. What is it?

Explanation / Answer

Given:

Height of the building = h = 6m

initial velocity of the 2nd ball= v1 =0

if t= time taken to reach the ground

then applying the equation of motion

h= v1t + 1/2 gt2

so, 6 = 1/2 x 9.8 x t2

so, t= 1.10 s

so time taken by the first ball to reach the ground = time taken by the 2nd ball + 2s

=1.1 + 2=3.1 s

Now applying the equation of motion

h= v1t + 1/2 gt2

so, 6 = v1 x 3.1 + 1/2 x 9.8 x 3.12

so, v1=-13.31 m/s

so the initial speed of the first ball must be 13.31 m/s so that both will hit the ground at the same time

b.

i. If the speed of the first ball is 13.0 m/s, then the height covered in t s

h=-13t + 1/2 x 9.8 x t2

similarly height covered by the 2nd ball in (t-2)s as it is dropped after 2s

h= 0 + 1/2 x 9.8 x (t-2)2

as both heights are equal

-13t + 1/2 x 9.8 x t2 = 1/2 x 9.8 x (t-2)2

so, 4.9 [ t2 - (t-2)2 ] =13t

so, 4.9 [ 4t -4]=13t

so, 19.6t-13t = 19.6

so, t = 2.969 s

substituting the same

h= 1/2 x 9.8 x (2.969-2)2

= 4.60 m

ii.

If the speed of the first ball is 19.2 m/s, then the height covered in t s

h=-19.2t + 1/2 x 9.8 x t2

similarly height covered by the 2nd ball in (t-2)s as it is dropped after 2s

h= 0 + 1/2 x 9.8 x (t-2)2

as both heights are equal

-19.2t + 1/2 x 9.8 x t2 = 1/2 x 9.8 x (t-2)2

so, 4.9 [ t2 - (t-2)2 ] =19.2t

so, 4.9 [ 4t -4]=19.2t

so, 19.6t-19.2t = 19.6

so, t = 49 s

substituting the same

h= 1/2 x 9.8 x (49-2)2

= 10824.1 m

c.As both cannot meet

  vmaxt + 1/2 gt2 > 1/2 x 9.8 x (t-2)2

so, vmaxt > 4.9 (-4t +4)

so, vmaxt > -19.6t +4

so, vmax > -19.6 +4/t As t>2

so, vmax > -17.6

so V0 should be more than 17.6 m/s

similarly

Vmin should be such that the first ball should reach the ground within 2 s

6 = Vmin x 2 + 1/2 x 9.8 x 22

so, Vmin =6.8 m/s

all the best in the course work

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