The vertical distance covered by a free falling object is given by the formula: S= ut+0.5at^2, where S is the distance covered (height of the building), u is the initial velocity (for this case it is 0 since the body is released from rest), t is the time taken for the object to hit the ground (it has taken 5 seconds) and a is the acceleration due to gravitational pull (assumed to be 9.8ms^2). Therefore, the height of the building is given by (0x5 +0.5x9.8 x25) which is 122.5m.
x = 1/2 G t2 = (1/2) (32.2) (9) = 144.9 ft (44.2 meters) (rounded)
54 m/s
The speed is 44.4... repeating metres per second.
Disregarding air resistance, what is the speed of a ball dropped from 12 feet just before it hits the ground? (Use 1 ft = 0.30 m, and use g = 9.8 m/s2.)
2 seconds. The time taken for the body to reach the ground does not depend on its mass.
1.56 seconds
It can take anywhere from just a few seconds, to several minutes, if the funnel cloud reaches the ground at all.
The speed is 44.4... repeating metres per second.
a. 144 feet b. 96 ft/sec.
Disregarding air resistance, what is the speed of a ball dropped from 12 feet just before it hits the ground? (Use 1 ft = 0.30 m, and use g = 9.8 m/s2.)
5 m
There is no reason for the object to change.
381 metres
4 seconds
44 meters tall
2 seconds. The time taken for the body to reach the ground does not depend on its mass.
1.56 seconds
176.4 meters
Ignoring air resistance . . .H = 1/2 G t2t = sqrt(2H/G) = sqrt(2 x 363 / 32.2) = 4.75 seconds (rounded)