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
-------------
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
A projectile is fired with an initial speed of 61.3m/s at an angle of 31.5° above the horizontal on a long flat firing range.
a)Determine the maximum height reached by the projectile.
b)Determine the total time in the air.
c)Determine the total horizontal distance covered (that is, the range).
d)Determine the speed of the projectile 1.44s after firing.
e)Determine the direction of the projectile 1.44s after firing.
1 answer