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.
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
... and by the way I am not the first Anonymous :)
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.
To find the maximum height reached by the bullet and the position where it hits the ground, we can use the equations of motion under constant acceleration.
First, let's analyze the motion in the vertical direction. The only force acting on the bullet vertically is the force due to gravity, which points downward with a magnitude of 9.8 m/s².
1. Maximum Height Reached:
The maximum height reached by the bullet occurs when its vertical velocity component becomes zero. We can use the kinematic equation to find the time it takes for the bullet to reach maximum height:
v_f = v_i + a*t
where
v_f = final velocity (0 m/s at the highest point),
v_i = initial velocity in the vertical direction (15 m/s upward),
a = acceleration due to gravity (-9.8 m/s²),
t = time taken to reach maximum height.
Rearranging the equation, we get:
t = (v_f - v_i) / a
Substituting the given values:
t = (0 - 15) / (-9.8)
t ≈ 1.53 seconds
Now, we can find the maximum height reached by the bullet using the kinematic equation:
h = v_i*t + (1/2)*a*t²
where
h = maximum height reached.
Substituting the values:
h = 15 * 1.53 + (1/2) * (-9.8) * (1.53)^2
h ≈ 23.09 meters
Therefore, the maximum height reached by the bullet is approximately 23.09 meters.
2. Position where it hits the ground:
To find the position where the bullet hits the ground, we need to determine the time it takes for the bullet to fall from the maximum height to the ground.
Using the equation:
h = v_i*t + (1/2)*a*t²
where
h = maximum height (23.09 meters),
v_i = initial vertical velocity (15 m/s upward),
a = acceleration due to gravity (-9.8 m/s²),
t = time taken to reach the ground.
Rearranging the equation, we get a quadratic equation in terms of t:
(1/2) * (-9.8) * t² + 15 * t - 23.09 = 0
Solving this quadratic equation, we find two possible values for t: t1 and t2.
Using the positive value of t (t1) since time cannot be negative, we can find the horizontal distance traveled by the bullet using the equation:
d = v_i * t
where
d = horizontal distance traveled.
Substituting the values:
d = 15 * t1
Finally, the position where the bullet hits the ground can be expressed as:
Position = initial position + horizontal distance traveled in the x-axis
Substituting the given initial position (9 i + 4 j + 15 k):
Position ≈ (9 + 15 * t1) i + 4 j + 15 k
Calculating the value of t1 using the quadratic formula, you can find the specific position where the bullet hits the ground.