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? 1 point25.4 𝘴17.3 𝘴43.0 𝘴33.6 𝘴
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? 1 point25.4 𝘴17.3 𝘴43.0 𝘴33.6 𝘴
Solution
To solve this problem, we need to use the equations of motion.
The height h of the shell at any time t can be given by the equation:
h = ut - 0.5gt^2
where: u is the initial velocity (225 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
We know that the shell is moving downward when it reaches a height of 6.20 x 10^2 m, which means it has already reached its maximum height and is now falling back down.
The time it takes for the shell to reach its maximum height can be found using the equation:
t = u/g
Substituting the given values:
t = 225 / 9.8 ≈ 22.96 s
This is the time it takes for the shell to reach its maximum height. However, we want to find the time it takes for the shell to reach a height of 6.20 x 10^2 m on its way down.
To find this, we can use the equation for h again, but this time we need to solve for t.
Substituting the given values:
6.20 x 10^2 = 225t - 0.5(9.8)t^2
Solving this quadratic equation, we get two solutions for t. The larger of the two will be the time at which the shell is at a height of 6.20 x 10^2 m and moving downward.
Solving the equation, we get t ≈ 43.0 s.
So, the correct answer is 43.0 s.
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