18. A cannon shell is fired straight up from the ground at an initial speed of 225 𝘮/𝘴. After how much time is the shell at a height of 6.20 3 102 𝘮 above the ground and moving downward?
Question
- A cannon shell is fired straight up from the ground at an initial speed of 225 𝘮/𝘴. After how much time is the shell at a height of 6.20 3 102 𝘮 above the ground and moving downward?
Solution
To solve this problem, we need to use the equations of motion. We know that the initial velocity (u) is 225 m/s, the final height (h) is 6.20 x 10^2 m, and we want to find the time (t) when the shell is moving downward.
The equation we will use is: h = ut + 0.5gt^2, where g is the acceleration due to gravity (-9.8 m/s^2 because it's acting downward).
First, we need to find the time (t1) it takes for the shell to reach its maximum height. We can use the equation v = u + gt, where v is the final velocity (0 m/s at maximum height).
0 = 225 + (-9.8)t1 => t1 = 225 / 9.8 ≈ 22.96 seconds
This is the time it takes for the shell to reach its maximum height.
Next, we need to find the time (t2) it takes for the shell to fall from its maximum height to the height of 6.20 x 10^2 m. We can use the equation h = 0.5gt^2 (since initial velocity is 0 at maximum height).
6.20 x 10^2 = 0.5 * 9.8 * t2^2 => t2^2 = (6.20 x 10^2) / (0.5 * 9.8) => t2 = sqrt((6.20 x 10^2) / (0.5 * 9.8)) ≈ 35.71 seconds
So, the shell is at a height of 6.20 x 10^2 m above the ground and moving downward after t1 + t2 = 22.96 + 35.71 ≈ 58.67 seconds.
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