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Grade: 12th pass
        
an object thrown vertically up from ground  passes the height 10 metre twice in the time interval of 10 sec the time of flight is
one month ago

Answers : (3)

Arun
22368 Points
							
Let the height of the highest point that the object reached be S meters
The object has climbed the height of (S-10) m in 10s/2 = 5 seconds and  it descended (S-10) m in the same time =  5sec
we use s = ut + 1/2 a t²
S – 10 = 0 * 5 + 1/2 * 9.8  * 5²
S = 132.5 meters
Time duration for the object to travel vertically 132.5 meters
132.5 = 0 t + 1/2 * 9.8 * t²
t² = 27.041      t = 5.196 sec
Total time of flight = 10.392 second
 
one month ago
Khimraj
3008 Points
							
Object passes through 10m height twice in 10s
so object reaches highest peak in 5s (as it will take equal time interval for going up and coming down) 
let velocity at highest point be V
applying equation of motion 
V=U+at
here V is final velocity
U is initial velocity
a is acceleration due to gravity(here it is negative acceleration as acceleration due to gravity acts in opposite direction, so a=-g=(-9.8m/s²)
here V=0, a=-9.8m/s², t=5s
0=U-9.8*5
U=49m/s
at 5m height V=49m/s, a=-9.8m/s²,s=5m
V²=U² + 2as
49²=U² - 2*9.8*10
U²=2597
U=51 m/s
Again V=U+a*t
so final velocity V=0, initial velocity U=51m/s, a=-9.8m/s², t we need to find
0=51-9.8*t
t=51/9.8
=5.2s (it is the time of going up)
So total time period is 2*5.2=10.4s
one month ago
aswanth nayak
66 Points
							
dear Student
 
 
The object has climbed the height of (S-10) m in 10s/2 = 5 seconds and  it descended (S-10) m in the same time =  5sec
we use s = ut + 1/2 a t²
S – 10 = 0 * 5 + 1/2 * 9.8  * 5²
S = 132.5 meters
Time duration for the object to travel vertically 132.5 meters
132.5 = 0 t + 1/2 * 9.8 * t²
t² = 27.041      t = 5.196 sec
Total time of flight = 10.392 second
 
regards
one month ago
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