Question

A stone thrown from the top of a building is given an initial velocity of 40.0 m/s straight upward. The building is 70.0m high, and the stone just misses the edge of the roof on its way down, determine
(A) The time at which the stone reaches its maximum height,
(B) The maximum height,
(C) The time at which the stone returns to the height from which it was thrown,
(D) The velocity of the stone at this instant
(E) The velocity and position of the stone at t = 10.00 s.

Answer 1

A. V = Vo + g*Tr.
0 = 40 - 9.8Tr, Tr = 4.08s = Rise time or time to reach max. ht.

B. hmax = ho + Vo*Tr + 0.5g*Tr^2.
hmax = 70 + 40*4.08 - 4.9*(4.08)^2 = 151.7 m. above gnd.

C. Tf = Tr = 4.08 s. = Fall time.

D. V = Vo + g*Tf = 0 + 9.8*4.08 = 40 m/s.

E. h = ho - 0.5g*Tf^2.
0 = 151.7 - 4.9Tf^2
4.9Tf^2 = 151.7,
Tf^2 = 31, Tf = 5.56 s. = Time to fall from max. ht. to gnd.

Tr+Tf = 4.08 + 5.56 = 9.64 s = Time in air. So the stone is on the gnd 10 s after release.

V^2 = Vo^2 + 2g*h = 0 + 19.8*151.7 = 3,003.7, V = 54.8 m/s.

Answer 2

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