X = 1/2 G t2
t = sqrt( 2X/G )
= sqrt ( 200/9.8 )
= sqrt (20.408 )
= 4.518 seconds (rounded)
1.56 seconds
Dropped objects hit the ground at the same time (as long as they're dropped from the same height) because the acceleration of gravity is constant. On earth, it's 9.8 meters per second (32.2 feet per second) every second.
We know that S=ut+.5gt*t here s=84m g=9.8m/s*s u=0 hence time is 4.14secs.
Assuming you have the same mass you could use the formula h=-16t^ 2+ c H stands for height of falling object after time c stands for height dropped from t stands for time
If the object that falls is sufficiently dense and streamlined, so that the effects of air resistance can be neglected, then you can estimate the height from which it fell like this: -- Time the fall, from the instant the object is released until the instant it hits the ground. -- The distance it fell is [ 16.1 T2 ] feet, or [ 4.9 T2 ] meters. The height of a tall building can easily be measured by using a barometer. There are two convenient methods. 1). Drop the barometer from the roof of the building. Time its fall, from the instant the barometer is released until the instant it hits the ground. The height of the building roof is [ 16.1 T2 ] feet, or [ 4.9 T2 ] meters. 2). Tie the barometer to the end of a long tape-measure or rope. With the barometer as a weight to resist the inaccuracy that would be introduced by the wind, slowly pay out the tape or the rope from the edge of the building's roof, until the barometer just touches the ground. At that point, either read the height directly from the tape, or mark the rope where it touches the roof, and measure the length of rope between the barometer and the mark later.
2s
4 seconds
It depends plus, you can't go to the top of the leaning tower of Pisa.
A pebble is dropped from the top of a 144-foot building. The height of the pebble h after t seconds is given by the equation h=−16t2+144 . How long after the pebble is dropped will it hit the ground?Interpretationa) Which variable represents the height of the pebble, and in what units?b) Which variable represents the time in the air, and in what units?c) What equation relates the height of the object to its time in the air?d) What type of equation is this?e) What are you asked to determine?
D = 1/2 g T2T = sqrt(2D/g) = sqrt( 109.2 / 9.8 ) = 3.335 seconds(rounded)
In a vacuum, just under 9 seconds. In the real world, more than that, but it depends on the shape of the object and the prevailing winds and air currents.
The ESB is much wider at its base than at its top, so no object dropped from its top would hit the sidewalk. HOWEVER, an object dropped from the height of the ESB would, if it experienced no air friction nor hit anything along the way, would hit the ground in 8.8 seconds. However, air friction would delay this by a few seconds, as a small ball would experience air resistance before that time.
Assuming they were in a vacuum, if both objects were dropped from th esame height, then both take the same length of time to reach the ground. All masses fall with the same acceleration, reach the same speed in the same period of time, and hit the ground at the same time. Otherwise and if there is an atmosphere or if they are dropped from different heights, you have not presented information; shape and size are the most important factors.
4 Seconds
1.56 seconds
a pool, a building, the ocean.
Dropped objects hit the ground at the same time (as long as they're dropped from the same height) because the acceleration of gravity is constant. On earth, it's 9.8 meters per second (32.2 feet per second) every second.