On flat ground, a cannonball is shot with a speed of v = 13.3 m/s over a wall ofheight h = 5 m located a distance d = 10 m away from the cannon. If xmax is the farthest away the ballcan be shot over the wall and xmin is the shortest distance the ball can reach while going over the wall,find xmax − xmin.
Question
On flat ground, a cannonball is shot with a speed of v = 13.3 m/s over a wall ofheight h = 5 m located a distance d = 10 m away from the cannon. If xmax is the farthest away the ballcan be shot over the wall and xmin is the shortest distance the ball can reach while going over the wall,find xmax − xmin.
Solution
To solve this problem, we need to use the equations of motion under constant acceleration (in this case, the acceleration due to gravity).
First, we need to find the time it takes for the cannonball to reach the height of the wall. We can use the equation:
h = vt - 0.5g*t^2
where: h = height of the wall = 5 m v = initial velocity = 13.3 m/s g = acceleration due to gravity = 9.8 m/s^2 t = time
Solving this equation for t gives two solutions, t1 and t2. We are interested in the smaller value (t1), which represents the time it takes for the cannonball to reach the height of the wall on its way up.
Next, we find the distances xmin and xmax, which are the distances the cannonball travels in the time t1 and t2 respectively. We can use the equation:
d = v*t
where: d = distance v = initial velocity = 13.3 m/s t = time
Finally, we find xmax - xmin by subtracting the two distances.
This problem involves quadratic equations and may require the use of a calculator or software to solve.
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