it would help if you typed your math so it can actually be read.
The Wikipedia article on "Trajectory" has several useful formulas, along with their derivations. Start there, and your way will be clear.
A cannon shoots a bullet from the roof of a building from the position (in SI units):
r i j k o
= 9 + 4 +15
and with initial velocity (SI units):
v i j k o
=13 + 22.5 +15
where the z-axis points upward in the vertical direction and with origin at the ground
level. Assuming that the air resistance is negligible, find the maximum height reached by
the bullet and the position where it hits the ground.
3 answers
r = 9 i + 4 j + 15 k in m
v = 13 i + 22.5 j + 15 k in m/s
now what I really have to do is solve the vertical problem first
Hinitial = 15 meters roof height
Vinitial = 15 m/s upward
v = Vi - 9.81 t assuming aa = -9.81 m/s^2 on earth
h = Hi + Vi t - 4.9 t^2
first top, when v = 0
0 = 15 - 9.81 t
t = 1.53 seconds upward
so
h at top = 15 + 15 * (1.53) - 4.9 * (1.53)^2
= 15 + 22.95 - 11.47
= 26.5 meters high above ground at top
now ground
when will h = 0 (ground)
4.9 t^2 - 15 t - 15 = 0
t = 3.855 seconds or -0.794, use positive time
NOW horizontal problem, goes at
u = 14 i + 22.5 j for 3.855 seconds so
x = 9 + 14 * 3.855
y = 4 + 22.5 * 3.855
v = 13 i + 22.5 j + 15 k in m/s
now what I really have to do is solve the vertical problem first
Hinitial = 15 meters roof height
Vinitial = 15 m/s upward
v = Vi - 9.81 t assuming aa = -9.81 m/s^2 on earth
h = Hi + Vi t - 4.9 t^2
first top, when v = 0
0 = 15 - 9.81 t
t = 1.53 seconds upward
so
h at top = 15 + 15 * (1.53) - 4.9 * (1.53)^2
= 15 + 22.95 - 11.47
= 26.5 meters high above ground at top
now ground
when will h = 0 (ground)
4.9 t^2 - 15 t - 15 = 0
t = 3.855 seconds or -0.794, use positive time
NOW horizontal problem, goes at
u = 14 i + 22.5 j for 3.855 seconds so
x = 9 + 14 * 3.855
y = 4 + 22.5 * 3.855
... and by the way I am not the first Anonymous :)
1 answer