Answer:
160 ft
Explanation:
As the canon ball was shot horizontally, its initial vertical velocity is 0. The ball vertical motion is generated by the gravitational acceleration g = 32m/s2. We can calculate the time it takes for it to drop 64 ft
[tex]s_v = gt^2/2[/tex]
[tex]t^2 = 2s/g = 2*64/32 = 4[/tex]
[tex]t = \sqrt{4} = 2s[/tex]
This is also the time it takes to travel horizontally at a constant rate of 80 ft/s if we ignore air resistance.
[tex]s_h = vt = 80 * 2 = 160 ft[/tex]
So the ball would land 160ft away from the wall on the ground
Answer:
159.57 Feet Horizontally from the foot of the wall
Explanation:
Given Data:
Height = [tex]h[/tex] = 64 feet
gravitational acceleration = g = 32.17 [tex]ft/s^{2}[/tex]
Initial Vertical Velocity component [tex]Vi = 0\\[/tex]
Initial Horizontal Velocity Component [tex]Ui[/tex] = 80 [tex]ft./s[/tex]
To find = Time taken to reach ground = [tex]t\\[/tex]
Horizontal Distance from the foot of the wall = [tex]x[/tex] = ?
Calculation:
From the equation of motion [tex]h = Vi*t + 1/2*g*t^{2}[/tex] ........ (1) (in terms of Vertical component of velocity).
Putting in values in the above equation.
[tex]t = 1.99 s[/tex] (Time taken to reach the ground)
Now, to find [tex]x[/tex]
we use the same equation of motion we used above but this time we solve for the horizontal component of Velocity
[tex]x = Ui*t + 1/2*a*t^{2}[/tex]
[tex]a = 0\\Ui = 80 ft/s\\t = 1.99 s[/tex]
Putting in values in the above equation.
[tex]x = 80*1.99\\x = 159.57 ft[/tex]
