A projectile is fired and follows a path described by the following
equations:
x(t) = (12.26 m/s)t
y(t) = 2.00 m + (10.28 m/s)t-(4.90 m/s2)t2
(a) What are the initial velocity components of the projectile?
vox
=
12.26 m/s
voy
=
10.28 m/s
(b) At the maximum height of the trajectory, what is the velocity
of the projectile?
v
=
vox
^
i
=
(12.26 m/s)
^
i
(c) At what angle was the projectile fired.
q
=
tan-1(voy/vox)
=
tan-1(10.28 m/s/12.26 m/s)
=
40.0o
(d) If the clock, originally set to zero when the projectile was
fired, reads 2.28 s when the projectile hits the ground, what is the
range of the projectile?
R
=
(12.26 m/s)(2.28 s)
=
28.0 m
(e) When the clock reads 1.05 s, the projectile is at its
maximum position above ground. Calculate this maximum height.
ymax
=
2.00 m + (10.28 m/s)(1.05 s)-(4.90 m/s2)(1.05 s)2
=
7.40 m
File translated from
TEX
by
TTH,
version 3.85. On 24 Sep 2015, 14:02.