Posted by **hamze** on Monday, April 1, 2013 at 7:09am.

A stone (on the edge of the roof) thrown from the top of

a building is given an initial velocity of 20.0 m/s straight

upward. The building is 50.0 m high, as shown in Fig. 4.

Using tA = 0 as the time the stone leaves the throwerâ€™s

hand at position ○A . The gravitational acceleration is 9.8

m/s

2

. 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 which the stone returns

to the height from which it was thrown;

(e) The velocity and position of the stone at t = 5.00s.

- civil engineering -
**bobpursley**, Monday, April 1, 2013 at 7:58am
a. at the top, vf=0

vf=vi-9.8t

b. hf=hi-1/2 g t^2

c. double time in a.

d. vf=vi-1/2 9.8 t

e.

- civil engineering -
**raymond**, Monday, November 23, 2015 at 2:43pm
a. 10.2 m/s t

b. 50469.8 ft

c. 104.04 v

d. 955.5 v

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