A shell is fired from the ground with an initial speed of 1.54 ✕ 103 m/s at an initial angle of 52° to the horizontal.
(a) Neglecting air resistance, find the shell's horizontal range.
(b) Find the amount of time the shell is in motion.
u = 1540 cos 52 = 948 the whole time
Vi = 1540 sin 52 = 1214 at start
v = Vi - 9.8 t
0 = 1214 - 9.8 t
t = 124 seconds to top
so
2 t = 248 seconds in the air (part B)
range = u * 2t = 234,871 meters or 235 km
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To find the shell's horizontal range and the time it is in motion, we can use the equations of projectile motion.
(a) The horizontal range of a projectile is given by the equation:
Range = (initial velocity * time of flight) * cos(angle)
Since the initial velocity and angle are given, we can calculate the horizontal range.
Range = (1.54 * 10^3 m/s * time of flight) * cos(52°)
To find the time of flight, we need to determine the time it takes for the shell to reach its highest point. The time of flight to reach the highest point is given by:
Time to reach highest point = (vertical component of initial velocity) / (acceleration due to gravity)
The vertical component of the initial velocity can be calculated as:
Vertical component of initial velocity = initial velocity * sin(angle)
Vertical component of initial velocity = (1.54 * 10^3 m/s) * sin(52°)
Now, we can calculate the time to reach the highest point:
Time to reach highest point = (1.54 * 10^3 m/s * sin(52°)) / 9.8 m/s^2
To find the total time of flight, we double the time to reach the highest point:
Total time of flight = 2 * time to reach highest point
Finally, substitute the value of the total time of flight back into the equation for range:
Range = (1.54 * 10^3 m/s * total time of flight) * cos(52°)
To find the horizontal range of the shell, we can use the equations of projectile motion.
First, we need to break the initial velocity into its horizontal and vertical components. The horizontal velocity (Vx) remains constant throughout the motion, while the vertical velocity (Vy) changes due to the effect of gravity.
The initial velocity (Vi) can be broken down as follows:
Vx = Vi * cosθ
Vy = Vi * sinθ
Given that the initial speed of the shell (Vi) is 1.54 × 10^3 m/s and the initial angle (θ) is 52°, we can calculate the horizontal and vertical components of the velocity:
Vx = (1.54 × 10^3 m/s) * cos(52°)
Vy = (1.54 × 10^3 m/s) * sin(52°)
Now, we can calculate the time of flight (t) using the equation:
t = 2 * Vy / g
where g is the acceleration due to gravity (9.8 m/s^2).
Next, we can find the horizontal range (R) using the equation:
R = Vx * t
Let's calculate both parts of the question step by step.
(a) Calculating the horizontal range (R):
Vx = (1.54 × 10^3 m/s) * cos(52°)
Vx ≈ 1.54 × 10^3 m/s * 0.614
Vx ≈ 945.56 m/s
t = 2 * Vy / g
Vy = (1.54 × 10^3 m/s) * sin(52°)
Vy ≈ 1.54 × 10^3 m/s * 0.793
Vy ≈ 1221.82 m/s
t = 2 * 1221.82 m/s / 9.8 m/s^2
t ≈ 249.3 s
R = Vx * t
R ≈ 945.56 m/s * 249.3 s
R ≈ 235,777.01 m or approximately 236 km
Therefore, the shell's horizontal range is approximately 236 km.
(b) Calculating the time of flight (t):
We have already calculated t in the previous step, which is approximately 249.3 s.
Therefore, the amount of time the shell is in motion is approximately 249.3 seconds.