Objects in free fall will be accelerating, so you need to know which second that you are interested in, and the acceleration from gravity (9.8 meters per sec2)
The formula for distance is: d = v0*t + (1/2)*a*t2. Where v0 is the initial velocity, t is time, and a is acceleration.
Impact velocity depends on the mass of the object and the height it falls from. It is the speed at which the acceleration due to gravity is maximized.
98 meters (322 feet) per second.
9.8 meters/second2 x 10 seconds = 98 meters/second.
2.
Approx 5.1 metres.
With the information given, all that can be said is that the distance is greater than the distance the object traveled in the previous second.
Impact velocity depends on the mass of the object and the height it falls from. It is the speed at which the acceleration due to gravity is maximized.
98 meters (322 feet) per second.
9.8 meters/second2 x 10 seconds = 98 meters/second.
No. Since the speed of a falling object keeps increasing, it falls through more distance in each second than it did in the second before.
2.
Approx 5.1 metres.
The speed stays thesame but the distance stays the same.
Assuming the object is falling near the surface of the Earth and neglecting air resistance, the object will fall approximately 4.9 meters in 1 second. This calculation is based on the acceleration due to gravity, which is approximately 9.8 meters per second squared.
Its acceleration due to gravity is constant. The acceleration is equal to the object's change in speed every second. I've tried to illustrate the constantly-increasing falling speed in my diagram below.
acceleration at surface on moon = 1.623 (m/s)/s. v = a*t = 1.623 * 1 = 1.623 metres / second
acceleration at surface on moon = 1.623 (m/s)/s. v = a*t = 1.623 * 1 = 1.623 metres / second