The object rises to the maximum height of 125 m in 5 second.
Also at the highest point the velocity of the ball must be zero, because if it is not zero then it can't be the heighest point.
Now by third equation of motion
Now taking upward direction as positive
The total displacement will be zero since the body comes back to the orginal position.
So total displacement=+125-125=0
but total distance=125+125=250m
Now we have to calculate the total time taken by the object in the journey.
By using second equation of motion,
where s=0 as total displacement is zero,
u=+50 as calculated
so putting these values in the equation we get
t=0 signifies that during the beginning of the journey also the net displacement was 0.t=10 is the time of completion of journey.
Average velocity=Total displacement/total time
Average speed=Total distance/total time
Hope it helps.
Distance = 138.28 m
Displacement = 88 m
distance=22m, half way =44/2=22
displacement=0, anything that moves in circular path will have 0 displacement
radius of the circle = 44m
half way. = 277/2=138.5m
Distance traveled = 2 × 2πr
= 2 × 2 × 22/7 × 5/2
= 31.43 m
displacement = 0 m
since initial point and final point are same.
by the distance traveled Whil going and coming back from school is 2km
for displacement if we take
West to East as x-axis
south to north as y-axis
so in x direction
= 23 i^
in y direction
= -22 j^
so net displacement
= 31.827 m
hope it helps you
Displacement = 31.8 m (approx)
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