A shell of a gun has a initial speed of 1700 m/s at initial inclination of 55◦ above the horiztontal. Ignore air resistance and other effects.
a) what is the maximum height?
B) what is the total time of the shell's flight?
c) how far away did the shell hit?
a) (Vo sin55)^2 = 2 g H
Solve for H.
b) g*(t/2) = Vo sin 55
Solve for t.
t = 2 Vo sin55/g
c) Range = 2(Vo^2/g)sin55*cos55
To solve this problem, we can use the equations of projectile motion. First, let's break down the information given:
Initial speed (u) = 1700 m/s
Initial inclination (θ) = 55° above the horizontal
Acceleration due to gravity (g) = 9.8 m/s² (assuming the shell is near Earth's surface)
Neglecting air resistance and other effects
a) What is the maximum height?
To find the maximum height, we need to determine the vertical component of the shell's initial velocity (v₀y) when it reaches its highest point. We can use the equation:
v = u + at
In the vertical direction, the final velocity (v) at the maximum height is zero since the shell momentarily comes to rest at that point. The acceleration (a) is the acceleration due to gravity, acting downward, and the initial velocity (u) is the vertical component of the initial velocity:
v = u + at
0 = v₀y + (-9.8)t
Rearranging the equation, we have:
t = v₀y/9.8
We can find v₀y using trigonometry. Since the inclination angle is given, we can use it to determine the vertical component of the initial velocity using sine:
v₀y = u * sin(θ)
Plugging in the values:
v₀y = 1700 * sin(55°)
v₀y ≈ 1395.28 m/s
Now we can find the time (t) it takes for the shell to reach its maximum height:
t = v₀y/9.8
t ≈ 1395.28/9.8 ≈ 142.20 seconds
The maximum height (h) can be calculated using the equation for vertical displacement:
h = v₀y² / (2g)
Plugging in the values:
h = (1395.28)² / (2 * 9.8)
h ≈ 98254.74 m ≈ 98.25 km
Therefore, the maximum height reached by the shell is approximately 98.25 kilometers.
b) What is the total time of the shell's flight?
The total time of the shell's flight is twice the time taken to reach the maximum height, as the time taken to ascend will be the same as the time taken to descend:
Total time of flight (T) = 2 * t ≈ 2 * 142.20 ≈ 284.40 seconds
Thus, the total time of the shell's flight is approximately 284.40 seconds.
c) How far away did the shell hit?
To determine the horizontal distance traveled by the shell, we can use the equation for horizontal displacement:
x = u * cos(θ) * t
Given that the initial inclination (θ) is 55° and the time (t) is 284.40 seconds (calculated in part b), we can find the horizontal distance:
x = 1700 * cos(55°) * 284.40
x ≈ 150461.60 m ≈ 150.46 km
Hence, the shell hit approximately 150.46 kilometers away.