A gunman standing on a sloping ground fires up the slope. The initial speed of the bullet is v0= 360 m/s. The slope has an angle α= 28° from the horizontal, and the gun points at an angle θ from the horizontal. The gravitational acceleration is g=10 m/s2. Let θ = 59°. Find the maximum range, lmax, of the bullet along the slope.
As you recall, the height of the bullet is given by
y(x) = tanθ x - g/(2(v cosθ)^2) x^2
and of course, the height of the slope is
y(x) = tanα x
So, using your numbers, you just need to find where the two curves intersect:
tan59° x - 5/(360 cos59°)^2 x^2 = tan28° x
I get x = 7787.2
Of course, that is the horizontal distance; I'm sure you can use that to find the distance up the slope ...
To find the maximum range (lmax) of the bullet along the slope, we need to use the projectile motion equations. Let's break down the problem into two components: horizontal and vertical.
Horizontal Component:
Since there is no acceleration in the horizontal direction, the only force acting on the bullet is the initial velocity (v0) in that direction. The horizontal component of the velocity (vx) remains constant throughout the motion. Therefore, we can use the following equation to find the time of flight (t) of the projectile:
t = lmax / vx
Vertical Component:
In the vertical direction, the bullet is subject to the force of gravity. Therefore, we need to consider the vertical component of the velocity (vy) and the initial angle (θ) with respect to the horizontal.
The initial vertical velocity (vy0) is given by:
vy0 = v0 * sin(θ)
The vertical displacement (hmax) reached by the bullet can be calculated using the equation:
hmax = vy0^2 / (2 * g)
At the maximum height, the vertical component of the velocity becomes zero, and the time taken to reach that height is half of the total time of flight (t/2).
The range along the slope (lmax) can be calculated as the horizontal displacement (lx) at the time (t/2) when the bullet returns to the ground level.
lx = vx * (t / 2)
Now let's substitute the given values into the equations and calculate the maximum range (lmax).
Given:
v0 = 360 m/s
α = 28°
θ = 59°
g = 10 m/s^2
Step 1: Calculate the horizontal component of the velocity (vx):
vx = v0 * cos(α)
Step 2: Calculate the time of flight (t):
t = lmax / vx
Step 3: Calculate the initial vertical velocity (vy0):
vy0 = v0 * sin(θ)
Step 4: Calculate the maximum height (hmax):
hmax = vy0^2 / (2 * g)
Step 5: Calculate the total time of flight (t):
t = 2 * (vy0 / g)
Step 6: Calculate the range along the slope (lmax):
lx = vx * (t / 2)
Substitute the values into the equations and calculate accordingly.