change in velocity is the same interval; thus the acceleration is constant. D FI
ID: 1525210 • Letter: C
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change in velocity is the same interval; thus the acceleration is constant. D FIGURE 2.26 Strobe photo of a freely falling ball. EXAMPLE 2.11 A ball on the roof Now let's examine the more complex case of a ball being thrown vertically upward. Even though the ball is initially moving upward, it is, nevertheless, in free fall and we can use Equations 2.12, 2.13, and 2.14 to analyze its motion. Suppose you throw a ball vertically upward from the flat roof of a tall building. The ball leaves your hand at a point even with the roof railing, with an upward velocity of 15.0 m/s. On its way back down, it just misses the railing. Find (a) the position and velocity of the ball 1.00 s and 4.00 s after it leaves your hand: (b) the velocity of the ball when it is 5.00 m above the railing; and (c) the maximum height reached and the time at which it is reached. Ignore the effects of the air.Explanation / Answer
When the ball goes to the top it has a velocity equal to zero. Then it starts accelerating downwards and by the time it reaches railing level it must have a downward velocity of 15m/sec( the same as the velocity with which it left your hand) , in order conserve energy.
The time taken for flight from hand to the maximum level where velocity of ball becomes zero, can be calculated using
v = u - gt
0= 15m/s - 9.8 t
solving ,we get t = 1.53 seconds ( this is time for upward journey of ball till it's velocity falls to zero)
In the upward journey the distance travelled can be calculated using
v^2 - u^2 = - 2g S
S = - 15^2/-2g = 11.48 meters ( from rail level to top most point of ball flight)
Now while coming down to ground level total distance to be travelled is = height of building + height from roof to peak point
S = 30m + 11.48 mrs = 41.48 mrs
To travel this distance let time taken be t '
The using s = ut + (1/2) g t ' ^2
41.48 mrs = 0+ 0.5 X 9.8 m/s^2 X t ' ^2
solving for t ' = 2.91 seconds
Hence total flight time from the time ball leaves hand till it hits ground= t + t ' = 1.53 s + 2.91 s = 4.44 secs
As already stated when ball reaches back to rail( hand) level its velocity will be 15 m/s in down-ward direction.
If V is velocity when ball hits ground,then
Hence,V^2 = (15m/s)^2 + 2 X 9.8 m/S^2 X 30 m = - 28.5 meters ( The root with negative sign is taken since in downward direction V will be negative )
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