Since the equation for time=sqrt(2h/g)
set 2 seconds for time and 9.8 for gravity
so 2=sqrt(2h/9.8)
clear the sqrt by squaring both sides
4= 2h/9.8
9.8*4 =2h
(9.8*4)/2 = height.
now that you have the height, you can do v=distance/time
v=height from the equation prior/2 seconds
i hope that works..
Formula: V(final) = V(initial) + at -a=acceleration=9.8m/ss, t=time Assuming the initial velocity is 0 (the rock was just dropped and not thrown)... 2 seconds - velocity = 19.6 m/s 5 seconds - velocity = 49 m/s 10 seconds - velocity = 98 m/s
Acceleration of gravity is 9.8 meters (32.2 feet) per second2.
After 2.5 seconds, the object has reached a speed of (2.5 x 9.8) = 24.5 m/s . . . . . (2.5 x 32.2) = 80.5 ft/sec
68.6 m/s.
initial velocity is the velocity with which a particle starts its journey.
The acceleration of gravity is 9.8 meters/sec2.In 1.5 seconds after it's dropped, any object is falling at (9.8 x 2) = 19.6 meters/sec.The mass of the rock makes no difference.
The displacement and velocity of a rock that is dropped from rest after 4s, is 6 km/h. This can vary depending on the speed of the rock, and the surroundings.
39 m\s downward
an object will fall at around 50ft every second, so around 3 seconds.
176.4 meters
49
speed is decided by v = a(t), where t is your time variable and a is your acceleration constant. after 3 seconds, your velocity v = (9.8m/s2)(3s) = 29.4m/s
29.4
49
initial velocity is the velocity with which a particle starts its journey.
The acceleration of gravity is 9.8 meters/sec2.In 1.5 seconds after it's dropped, any object is falling at (9.8 x 2) = 19.6 meters/sec.The mass of the rock makes no difference.
The displacement and velocity of a rock that is dropped from rest after 4s, is 6 km/h. This can vary depending on the speed of the rock, and the surroundings.
80 ft / sec ~ 55mph The above ignores air resistance which will not be significant at this speed.
S=ut+0.5*a*t2 s=displacement u=intial velocity=0m/s t=time=7seconds a=9.81m/s2 s=0*7+0.5*9.81*72 s=240.345m
,033
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.