1)Consider the same situation, but now let the initial speed V^0 of the first ba

Question

1)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^0 = 8.60 .

2)If V^0 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 .

3)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

1) t1 - t2 = 1.16 s, v0 = 8.60 m/s, find h

displacement of the first ball = -h = v0t1 - gt12/2,

solve for t1 and get t1 = [v0 + (v02 + 2gh)]/g

displacement of the second ball = -h = -gt22/2,

t2 = (2h/g)

[v0 + (v02 + 2gh)]/g - (2h/g) = 1.16

[v0 + (v02 + 2gh)] - (2gh) = 1.16g

let u = (2gh)

v0 + (v02 + u2) - u = 1.16g

(v02 + u2) = u + 1.16g - v0

v02 + u2 = (u + 1.16g - v0)2

v02 + u2 = u2 + (1.16g - v0)2 + 2(1.16g - v0)u

0 = (1.16g)2 - 2*1.16g*v0 + 2(1.16g - v0)u

u = 1.16g (2*v0- 1.16g)/[2(1.16g - v0)] = 11.9758 = (2gh)

h = 7.317 m

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