The impulse that Earth exerts on the apple is equal to the change in momentum of the apple. We can use the formula:
impulse = force × time = mass × acceleration × time
The acceleration due to gravity is approximately 9.8 m/s^2. We can assume that the apple starts from rest, so its initial momentum is zero. Therefore, the impulse during the first 0.50 m of its fall is:
impulse = force × time = (0.150 kg) × (9.8 m/s^2) × (0.50 s) = 0.735 N·s
During the next 0.50 m of its fall, the apple will have picked up some speed, so its momentum will increase. Let's assume that its final velocity after falling 0.50 m is v. We can use the equation:
v^2 = u^2 + 2as
where u is the initial velocity (which is zero), a is the acceleration due to gravity, and s is the distance fallen (which is 0.50 m). Solving for v, we get:
v = sqrt(2as) = sqrt(2 × 9.8 m/s^2 × 0.50 m) = 3.13 m/s
The change in momentum during the second 0.50 m of its fall is:
Δp = mvf - mvi = (0.150 kg) × (3.13 m/s) - (0.150 kg) × (0 m/s) = 0.47 kg·m/s
The impulse during the second 0.50 m of its fall is also 0.735 N·s, since the force of gravity is constant. So the total impulse during the first meter of its fall is:
impulse = 0.735 N·s + 0.735 N·s = 1.47 N·s