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 1.20 s later. You may ignore air resistance.

A) If the height of the building is 20.7 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 = 8.65 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.

Answer 1

A) it is given that at some point both the ball hit the ground at the same time

here we have

s=ut+1/2at².................1)

here a=-g ( since the ball is going down)

here the u=0 since its starting from rest

-20.7=-1/2*9.8*t²

t=2.0554

the second ball is dropped from the roof 1.2 s later so total time is t=2.0554+1.20=3.2554s

now again using eqn 1)

-20.7=u*3.2554-1/2*9.8*3.2554²

u=9.59m/s

so the answer is 9.59m/s or 9.60 m/s

B) here we have

-y=-1/2*9.8*(t-1.2)²...........1)

-y=8.65m/s*(t)-1/2*9.8*(t)².........2)

solving 1 and 2 we get,

8.65*t-4.9t²=-4.9t²+11.76t-7.056

(11.76-8.65)t=7.056

3.11t=7.056

t=2.2688s

now putting the value of t in eqn 1 we have

y=1/2*9.8*(2.2688-1)²

y=7.89 m

so the answer is 7.89 m or 7.90 m

C) the maximum velocity will be the time difference between the two balls to land under acceleration due to gravity

Vmax=g*t=9.8m/s²*1.2s=11.8 m/s

so the answer is 11.8 m/s

D) and V min will be the half of the maximum velocity

Vmin=Vmax/2=11.8/2=5.88m/s

so the answer is 5.88 m/s or 5.90 m/s