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 1.06 s later. You may ignore air resistance.

A. If the height of the building is 19.4 m , what must the initial speed be of the first ball if both are to hit the ground at the same time? (already found this answer to be 8.58 m/s)

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 = 8.70 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

a)

Let the speed of first ball be v
The height, H which it reaches above the building = (v^2/2g) = H
The time, T to go to the top for ball 1 = v/g = T
Let the height of the building be h. we have
v/g + sq rt [2*(H+h)/g] = 1.06 + sq rt[2h/g] = 1.06 + sq rt[(2*19.4)/9.8} = 3.04, Multiplying by g gives,
sq rt [2*g*(H+h)] =3.04 *9.8 - v , Squaring we get
2gH + 2gh = (29.87 -v)^2 or
v^2 +2gh = (29.87 -v)^2 or
2gh = (29.87 - v)^2 - v^2 = 29.87*(29.87 - 2*v) = 2*9.8*19.4 = 380.24 or
59.74*v = 511.97

v = 8.56 m/s

2. Same equations will hold and derivations will work; v will be repalced by Vo = 8.70
Similarly we have,
Vo/g + sq rt [2*(H+h)/g] = 1.06 + sq rt[2h/g]; multiplying by g we get
Vo + sq rt[2*g*(H+h)] = 1.06*g + sq rt[2gh]
we are suppoed to find out h, but 2gh = V^2,say then if we know V, the velocity with which the second ball reaches ground we can know h. replacing variable h by vaiable V and putting all data, we have
8.70 + sq rt[Vo^2+V^2] = 1.06*9.8 + V = 10.38 +V or
sq rt[Vo^2+V^2] = 1.68+ V or
Vo^2 +V^2 = 1.68^2 +V^2 + 2*1.68*V or
8.70^2 - 1.68^2 = 79.45*V = 72.86 or V = .91m/s
So h = (.91^2)/(2*9.8)
= 0.04 m

3 and 4
Consider the following general relation in 2.
Vo^2 +V^2 = 2.67^2 +V^2 + 2*2.67*V or
Vo^2 = 7.13 +5.34*V
Now V cannot be negative So V0^2 > or = 7.13
so Vomin = sq rt(7.13) = 2.67
For Vo max consider such value of V0 so tthat the up and down trip is completed in 1.06 s. Then teh secnd ball can never reach in same time as first.
So Vo max is given by
[(2*Vomax)/g] must not be greater than or = 1.06 s or
Vomax = (9.8*1.06)/2 = 5.19 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.