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A cannon, placed in the valley, fires a shell at 50° with respect to the horizontal line. The initial speed of the shell is 700 m/s.

a) Find the time it takes the shell to hit the ground.
(b) Calculate the range of the shell.

User Kst
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Final answer:

The time it takes for the shell to hit the ground is approximately 92.4 seconds. The range of the shell is approximately 41.6 kilometers.

Step-by-step explanation:

To find the time it takes for the shell to hit the ground, we can use the fact that the vertical motion of the shell is affected only by the gravitational force. We know that the initial angle of the shell is 50° with respect to the horizontal line, so we need to break down the initial velocity into its horizontal and vertical components. The initial vertical velocity can be found using the formula vₒy = vₒsinθ, where vₒ is the initial speed and θ is the initial angle. In this case, vₒ = 700 m/s and θ = 50°, so vₒy = 700 m/s * sin(50°) = 450.16 m/s. We can use the formula s = vₒt - (1/2)gt² to find the time it takes for the shell to hit the ground. Setting s = 0 (since the shell hits the ground), vₒ = vₒy, and g = 9.8 m/s², we get 0 = (700 m/s * sin(50°))t - (1/2)(9.8 m/s²)t². Solving this quadratic equation gives us t = 92.4 seconds.

To calculate the range of the shell, we need to find the horizontal distance traveled by the shell. The horizontal velocity of the shell remains constant throughout its motion and can be found using the formula vₒx = vₒcosθ, where vₒ is the initial speed and θ is the initial angle. In this case, vₒ = 700 m/s and θ = 50°, so vₒx = 700 m/s * cos(50°) = 449.69 m/s. To find the range, we can use the formula R = vₒx * t, where R is the range and t is the time it takes for the shell to hit the ground. Plugging in the values, we get R = 449.69 m/s * 92.4 s = 41,569.24 meters or approximately 41.6 kilometers.

User Rjv
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