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A shell is fired from the ground with an initial speed of 1.7x10^3 m/s at an angle of 55 degrees to the horizontal. what is the amount of time the shell is in motion

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Answer:

To find the amount of time the shell is in motion, we can analyze the motion of the shell separately in the horizontal and vertical directions.

Given:

Initial speed (u) = 1.7 × 10^3 m/s

Launch angle (θ) = 55 degrees

Horizontal Motion:

The horizontal motion of the shell is unaffected by gravity. The horizontal velocity (vx) remains constant throughout the motion.

vx = u * cos(θ)

Vertical Motion:

The vertical motion of the shell is influenced by gravity. We can analyze it using the equations of motion.

Initial vertical velocity (uy) = u * sin(θ)

Acceleration due to gravity (g) = 9.8 m/s^2

Using the equation of motion:

y = uy * t - (1/2) * g * t^2

At the highest point of the trajectory, the vertical velocity becomes zero. We can find the time it takes to reach the highest point using the equation:

0 = uy - g * t_max

t_max = uy / g

The total flight time will be twice the time it takes to reach the highest point (as the shell takes the same amount of time to reach the highest point as it does to fall back to the ground):

Total flight time = 2 * t_max

Let's calculate the time:

Substituting the given values:

u = 1.7 × 10^3 m/s

θ = 55 degrees

g = 9.8 m/s^2

uy = u * sin(θ)

t_max = uy / g

Total flight time = 2 * t_max

Calculating:

uy = (1.7 × 10^3 m/s) * sin(55 degrees)

uy ≈ 1.7 × 10^3 m/s * 0.819

uy ≈ 1392.3 m/s

t_max = (1392.3 m/s) / (9.8 m/s^2)

t_max ≈ 142.04 seconds

Total flight time = 2 * 142.04 seconds

Total flight time ≈ 284.08 seconds

Therefore, the shell is in motion for approximately 284.08 seconds.

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User Costa Walcott
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