Final answer:
To find the magnitude of the acceleration of the shell just before it strikes the ground, you can use the equations of motion. The acceleration in the x-direction is zero since it is fired horizontally, and in the y-direction, you can use the equation: y = y0 + v0y t + (1/2)at^2. Solving for a, you can find the magnitude of the acceleration.
Step-by-step explanation:
Since the shell strikes the ground, its final position y is equal to the height of the cliff, which is 80 m. The initial position y0 is 0 m. The initial vertical velocity v0y is also 0 m/s because the shell is fired horizontally. Solving for a, we get:
a = (2(y - y0))/(t^2)
Now we need to find the time it takes for the shell to hit the ground. We can use the equation:
x = v0x t
where x is the horizontal displacement, v0x is the initial horizontal velocity, and t is the time. Since the shell strikes the ground 1330 m from the base of the cliff, we have:
v0x t = 1330
Since the initial horizontal velocity v0x is constant, we can solve for t:
t = 1330/v0x
Now we can substitute this value of t into the equation for acceleration:
a = (2(y - y0))/((1330/v0x)^2)
Finally, we can calculate the magnitude of the acceleration by taking the absolute value of a:
|a| = |(2(y - y0))/((1330/v0x)^2)|