Find the angle of elevation of a gun which fire a shell with velocity 120m/s and hit a target on the same level but at 1300m
u = 120 cos T forever
Vi = 120 sin T initially
so
Range = u t = 1300
t = 1300/(120 cos T)
h = 0 + Vi t - 4.9 t^2
0 = Vi t - 4.9 t^2
t = Vi/4.9
so
1300/(120 cos T) = 120 sin T / 4.9
now it is just algebra and trig
remember
cos T sin T = (1/2) sin 2T
To find the angle of elevation of a gun, we can use the concept of projectile motion.
Let's start by breaking down the problem into two components: the horizontal and vertical motion of the shell.
1. Vertical Motion:
The vertical motion can be described using the equation of motion for vertical motion:
y = ut + (1/2)gt^2,
where:
y = vertical displacement (height above the ground, in this case)
u = initial vertical velocity (0 m/s, since the shell starts from the same level)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach the target (which is the same for horizontal and vertical motion)
In this case, we want to find the time taken for the shell to reach a height of 1300m. So we can rewrite the equation and solve for time (t):
1300 = (1/2)(-9.8)t^2
Solving for t, we get:
t^2 = (1300 * 2) / -9.8
t^2 = 265.31
t ≈ √265.31
t ≈ 16.28 seconds
2. Horizontal Motion:
The horizontal motion of the shell is given by the equation:
x = ut,
where:
x = horizontal displacement (range)
u = initial horizontal velocity (which is the same as the initial velocity of the shell, 120 m/s)
t = time taken to reach the target
We know that the horizontal distance traveled (x) is 1300m, and the time (t) is 16.28 seconds (from the vertical motion calculation). So we can plug in the values and find the initial velocity (u):
1300 = 120 * 16.28
Solving for u, we get:
u = 1300 / 16.28
u ≈ 79.83 m/s
Now that we have the initial vertical and horizontal velocities (0 m/s and 79.83 m/s, respectively), we can find the angle of elevation (θ) using trigonometry.
Using the formula:
tan(θ) = vertical velocity / horizontal velocity
We can substitute the values:
tan(θ) = 79.83 / 120
Solving for θ, we take the inverse tangent:
θ ≈ tan^(-1)(79.83 / 120)
θ ≈ 37.6 degrees
Therefore, the angle of elevation of the gun would be approximately 37.6 degrees.