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=vi9.8t
b. hf=hi1/2 g t^2
c. double time in a.
d. vf=vi1/2 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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