When the object is about to hit the ground, its vertical velocity will be negative (downward) due to the effect of gravity. To find the acceleration just before it hits the ground, we can use the equation:
v^2 = u^2 + 2as
Where: v is the final velocity (which is the velocity just before hitting the ground) u is the initial velocity (80 m/s in this case) a is the acceleration s is the displacement (which is the negative of the initial height)
We want to find the acceleration, so we rearrange the equation:
a = (v^2 - u^2) / (2s)
Given that the object reaches a maximum height of approximately -326.53 meters (as calculated in the previous response), the displacement (s) would be -326.53 meters.
Now, we know that the final velocity just before hitting the ground is the negative of the initial velocity (-80 m/s).
Substituting the values into the equation:
a = ((-80)^2 - 80^2) / (2 * (-326.53)) a = (6400 - 6400) / (-653.06) a = 0 m/s^2
The acceleration just before the object hits the ground is 0 m/s^2. This means that just before impact, the object is not accelerating vertically; its velocity is constant.