Let v be the velocity when the ball is at 640 feets going downwards v = 48 feet /sec let the velocity with which it reaches the ground be u then, u2=v2+2gh g = acc due t ogravity in feet/sq.sec h = 640 feet the time taken to reach the ground = time to return to 640ft + the time to fall from there Time taken to get to the ground is 8 seconds. Final velocity is 208 feet per second downwards
The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2
If the engineer's eye is at ground level, then the distance to the point on the building underneath its highest point is 450/tan(22) ft. If the engineer was standing and his eyes were x ft above the ground, the distance is (450-x)/tan(22) ft.
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
8.5 + 2.75 = 11.25
31 m/s
mgh = 1/2 * m * v^2 v = sqrt (2 * 9.8 * 1.2) v = 4.8 m/s [down]
529.2 J
The ball had potential energy before it was dropped. This potential energy was due to its position above the ground.
No because you touch yourself at night.|_ 4 VV |_Just kidding. It does!Psyche!Actually, it does not because when your mother dropped you from 2 meters above the ground, you fell at a negative VELOCITY. Speed does not specify direction, and therefore can not be negative.
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
My contractor suggested I invest in an above ground pool deck rather than the usual. What is the benefit?
lofty means high, so a lofty building is perhaps a building built high above the ground or maybe a skyscraper?
The vertical component of velocity for the projectile when it is fired horizontally is zero. This is because the initial velocity is entirely in the horizontal direction, and there is no initial velocity in the vertical direction. Gravity will act on the projectile, causing its vertical velocity to increase as it travels.
According to Building Control Regulations, the DPC should be a minimum of 150mm above external finished ground level.
Assume that acceleration due to gravity, g = 9.8 ms-2. v2 = u2 + 2gs u, the initial velocity is 0 ms-1, s, the distance travelled is 140 - 20 = 120 m. So v2 = 2352 m2 so that v = 48.5 ms-1 approx.
A cellar (or basement) is below the ground level of a house or building.