If the ball was dropped from a roof and hit the ground 3.03 seconds later, then when it hit the ground
its velocity was 29.694 meters (97.42 feet) per second (rounded) downward.
Ignoring air resistance, the velocity of any object that goes off a cliff is 29.4 meters (96.5 feet) per second downward, after 3 seconds in free-fall.
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant before it hits the ground?
Assuming you throw the rock horizontally off the cliff it drops down at the acceletrtion of gravity. height= 1/2 gt^2 With g = 9.8 m/sec and t = 5 seconds we have height = (1/2) (9.8)(5)(5) = 122.5 meters notice it has nothing todo with the 50 meter distance, which depends on the horizontal velocity.
Distance of fall in T seconds = 1/2 g T2Distance of fall in 2 seconds = (1/2) (9.8) (2)2 = (4.9 x 4) = 19.6 metersHeight of this particular ball after 2 seconds = (70 - 19.6) = 50.4 meters
7
Ignoring air resistance, the velocity of any object that goes off a cliff is 29.4 meters (96.5 feet) per second downward, after 3 seconds in free-fall.
18
593 meters.
2.76 meters/second
A .50 magnum has a muzzle velocity of about 420 meters per second (1480 feet per second). The time for a .50 caliber bullet to travel one mile at a constant speed of 1400feet per second is around four seconds. After four seconds, the bullet will have dropped around 256 feet. In reality, the bullet is slowed down due to air resistance, and takes longer to travel one mile (thus it drops further). Also note that in theory, the caliber of the bullet does not determine how far it drops after traveling one mile. The muzzle velocity is the key parameter in determining this.
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant before it hits the ground?
The initial acceleration will be about 9.8 meters per second squared (m/s2) and the acceleration gradually drops to zero due to air resistance. The terminal velocity varies greatly for a person curled up into a ball and a person with a parachute.
Well, to answer this you need to find the meters per second. To find that we need to see how many meters the hammer moves in one second. To do this just divide the meters traveled by the seconds taken: 6/2.7 = 2.2222 (repeating) don't be frightened by the repeating 2s this just means that it's a fraction. The number can be changed into: a mixed number: 2 2/9 (That's 2 and 2 over 9) or as a mixed number 20/9 (20 over 9)
Assuming you throw the rock horizontally off the cliff it drops down at the acceletrtion of gravity. height= 1/2 gt^2 With g = 9.8 m/sec and t = 5 seconds we have height = (1/2) (9.8)(5)(5) = 122.5 meters notice it has nothing todo with the 50 meter distance, which depends on the horizontal velocity.
The duration of Eye Drops is 1800.0 seconds.
Distance of fall in T seconds = 1/2 g T2Distance of fall in 2 seconds = (1/2) (9.8) (2)2 = (4.9 x 4) = 19.6 metersHeight of this particular ball after 2 seconds = (70 - 19.6) = 50.4 meters
Use the formula for constant acceleration; in the simplest case, where the initial velocity is zero, it is simply:distance = (1/2) times acceleration x time squared You can use 9.8 meters/second squared for acceleration; the distance will then be in meters.