Question 1207029: Example on free fall:
1. A body is released from rest and falls freely. Compute its position and velocity after 1 and 2s. Take the origin at the elevation of the starting point, the y-axis vertical, and the upward direction as positive.
2. A stone is thrown from the top of a building with an initial velocity of 20 m/s straight upward. The building is 50 m high, and the stone just misses the edge of the roof on its way down. Determine:
(a) the time needed for the stone to reach its maximum height,
(b) the maximum height,
(c) the time needed for the stone to return to the level of thrower,
(d) the velocity of the stone at this instant, and
(e) the velocity and the position of the stone at t =5s.
(f) the velocity of the stone just before it hits the ground
Answer by mananth(16946)   (Show Source):
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1. A body is released from rest and falls freely. Compute its position and velocity after 1 and 2s. Take the origin at the elevation of the starting point, the y-axis vertical, and the upward direction as positive.
An object that is moving under only the influence of gravity is in free fall
Acceleration due to gravity is generally taken as 9.8 m/s^2
We use newton's equation of motion
v=u+at
v = final velocity
u initial velocity, a =Acceleration due to gravity, t = time in seconds
The object falls from rest so u=0 a= g = -9.8 m/s^2, t = 1 (s)
v= 0+(-9.8)(1) = -9.8 m/s
Distance = average velocity * time = (0+(-9.8))/2 *1= -4.9 *1 = -4.9 m (After 1 second)
Similarly find for 2 seconds here t=2
2. A stone is thrown from the top of a building with an initial velocity of 20 m/s straight upward. The building is 50 m high, and the stone just misses the edge of the roof on its way down. Determine:
(a) the time needed for the stone to reach its maximum height,
Newton's First law of motion
v= u+at
0= 20+(-9.8) *t (The object reaches maximum height) v=0
-20/-9.8 = t
t = 2.04 s
(b) the maximum height,
Newton.s second law of motion
s= ut+1/2 g t^2 ( here a=g)
s = 20*(2.04) + 1/2 * (-9.8)*(2.04)^2
s= 20.40 m
(c) the time needed for the stone to return to the level of thrower,
time to go up = time to come down
2*2.04 = 4.08 s
(d) the velocity of the stone at this instant, and
The object is falling under gravitational force only
t = 2.04 s ,g= -9.8 m/s^2
v= 0+2.04*(-9.8)
v = -19.992 m/s
(e) the velocity and the position of the stone at t =5s.
Again v= u+at
v = 0+(-9.8)*5
v = -49m/s
Position
v^2= u^2 +2gs
-49^2= 0^2+ 2*(-9.8)*s ( s is distance)
(-49)^2/(2*(-9.8) )= s
s= -122.5 m
(f) the velocity when it touches the ground
The height from which it is falling is 50 (height of building)+20.40( maximum height reached) = 70.40 m
v^2= u^2+2gs ( Newton's Third law of motion)
v^2 = 0+2*(-9.8)*70.40
v= sqrt(2*(-9.8)*70.40)
v = -37.15 m/s
The negative signs indicate downward direction
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