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 1.16 s later. You may ignore air resistance.

a. If the height of the building is 19.0 m , what must the initial speed be of the first ball if both are to hit the ground at the same time?

b. Consider the same situation, but now let the initial speed v0 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 v0 = 9.00 m/s .

c. If v0 is greater than some value vmax, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for vmax.

d. If v0 is less than some value vmin, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for vmin.

Explanation / Answer

part A :

apply for second, -0.5 gt^2 = -H

0.5 * 9.8 *t^2 = 19

t = 1.96 secs

now for first ball, again (0.5 g (t + 1.16)^2 = -19

Vo = 9.25 m/s

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part B :

dY1 = dY2

-0.5 g t^2 = -0.5 g (1+ 1.16)^2 + 9*(t +1.16)

t = 1.62 secs

Height H = 0.5 * 9.8 * 162^2

H = 12.9m or 13 m

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Part C:

if H = 0

t = 2Vo/g

vo = gt/2 = (9.8 * (1.16+1.62)/2

Vmax = 13.66 m/s

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part D :

if H = 0

t = 2Vo/g

vo = gt/2 = (9.8 * (1.16)/2

Vmin = 5.684 m/

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