Acceleration of gravity = 9.8 m/sec2.
39.2 / 9.8 = 4 seconds
Ignoring any effects due to air resistance, the speed of the stone is zero at the instant it's dropped, and increases steadily to 78.98 meters per second when it hits the ground. The velocity is directed downward throughout the experiment.
You have to let me ignore air-resistance for this one. Since I don't know anything about the shape or size of the stone, I can't calculate any effects of air. Initial velocity = 0 Final velocity = 5 x acceleration = 5 x G = 5 x (9.8 meters / sec2) = 49 meters per second Average velocity during the fall = 1/2 x ( 0 + 49 ) = 24.5 meters per second Distance of the fall = (24.5 meters per second) x (5 seconds) = 122.5 meters = 401.9 ft(approx., rounded)
Acceleration is the rate of change of velocity - how fast a velocity changes. Therefore, its units are naturally (meters/second) / second, usually written as meters/second2.Acceleration is the rate of change of velocity - how fast a velocity changes. Therefore, its units are naturally (meters/second) / second, usually written as meters/second2.Acceleration is the rate of change of velocity - how fast a velocity changes. Therefore, its units are naturally (meters/second) / second, usually written as meters/second2.Acceleration is the rate of change of velocity - how fast a velocity changes. Therefore, its units are naturally (meters/second) / second, usually written as meters/second2.
Assuming that the gun is fired parallel to the ground, the bullet will begin to fall the instant that it leaves the muzzle. The total fall will be 200 meters. You will need to calculate how long it will take an object to fall 200 meters (hint- about 9.753 meters per second per second- or 9.753 meters the first second, 19.50 meters during the second second, etc) THEN multiply the velocity of the bullet (643 meters/ second) by the number of seconds it is in flight. That will be the distance when it hits the ground.
An acceleration is not a velocity - it is the rate of change of velocity. In SI units, the units of velocity are meters/second. Acceleration is the rate of change of velocity, per unit time - how fast the velocity changes. Therefore, its units are velocity / time. In SI units, this gives you (meters/second) / second, usually written as meters/second2.
19.6m
Ignoring any effects due to air resistance, the speed of the stone is zero at the instant it's dropped, and increases steadily to 78.98 meters per second when it hits the ground. The velocity is directed downward throughout the experiment.
20 meters per second
Answer: 44 meters
44 meters tall
Acceleration = (change in velocity) / (time for the change)9.8 = (change in velocity) / (2 seconds)9.8 x 2 = change in velocity = 19.6 meters per second .Hint: The mass of the object and the height of the building are there just tothrow you off balance. You don't need either of them to answer the question.
The building is h=.5 gt^2 meters tall; that is = .5x9.8 x25 =122.5 meters.
In two seconds of fall, the speed increases 19.6 meters (64.4 feet) per second. The magnitude of velocity increases by that amount, while the direction of velocity doesn't change.
You have to let me ignore air-resistance for this one. Since I don't know anything about the shape or size of the stone, I can't calculate any effects of air. Initial velocity = 0 Final velocity = 5 x acceleration = 5 x G = 5 x (9.8 meters / sec2) = 49 meters per second Average velocity during the fall = 1/2 x ( 0 + 49 ) = 24.5 meters per second Distance of the fall = (24.5 meters per second) x (5 seconds) = 122.5 meters = 401.9 ft(approx., rounded)
If the ball was dropped from a roof and hit the ground 3.03 seconds later, then when it hit the groundits velocity was 29.694 meters (97.42 feet) per second (rounded) downward.
319 meters = 1046 feet high 77 floors above ground.
Ther velocity when falling 1000 meters is v=sqroot(2x1000x9.8) = 140 meters/second.