Posted by juanita on Saturday, November 24, 2012 at 8:19pm.
a
vertical problem
vi = 61.3 sin 31.5 = 32 m/s up
when at top (v = 0) ?
o = 32 - 9.81 t
t = 3.26 seconds to max ht
how high?
h = Vi t - (1/2)(9.81)t^2
h = 32 (3.26) - 4.9(10.7)
= 51.9 m high
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easier way by the way:
average speed up = 32/2 = 16m/s for 3.26 seconds --> 52 m
--------------------------
b
time in air = twice the time up = 3.26*2 = 6.52 seconds
c
horizontal problem
constant horizontal speed = 61.3 cos 31.5 = 52.3 m/s
time is still 6.52 s
range = 52.3 * 6.52 = 341 meters
d
vertical problem
v = 32 - 9.81 (1.44)
= 17.9 m/s
horizontal speed is always 52.3
s = sqrt(17.9^2+2.3)^2
e
tan angle up from hor = 17.9/52.3
= 18.9 degrees up from horizontal
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