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A stone is projected with a velocity 20√2 m/s at an angle of 45° to the horizontal .The average velocity of stone during the motion from starting point to its maximum height is

A stone is projected with a velocity 20√2 m/s at an angle of 45° to the horizontal .The average velocity of stone during the motion from starting point to its maximum height is 

Grade:12th pass

3 Answers

Khimraj
3007 Points
5 years ago
average velocity = (total displacement)/(time taken)
max height = u2cos245/2*g = 20 m
Half range = u2sin2*45/2*g = 40 m
time taken = 2usin45/g = 4 sec
total displacement = \sqrt{20^{2} + 40^{2}} = 20\sqrt{5}
so average velovity = 20\sqrt{5}/4 = 5\sqrt{5} m/s
Hope it clears. If you like answer then please approve it.
naman
30 Points
5 years ago
 
average velocity = (total displacement)/(time taken)
max height = u2cos245/2*g = 20 m
Half range = u2sin2*45/2*g = 40 m
Time half the time of flight that is 2 seconds
Average velocity equals to total displacement upon total time therefore average velocity equals to 10√5
SANJAY MODI
13 Points
5 years ago
Initial velocity:
\vec{u}=ucos45^o\hat{i}+usin45^0\hat{j}=20\hat{i}+20\hat{j}m/s
Final velocity:
\vec{v}=ucos45^{o}\hat{i}=20\hat{i}m/s
Therefore average velocity:
\vec{v_{av}}=\frac{\vec{u}+\vec{v}}{2} 
as the motion is uniformly accelerated:
Putting values:
\vec{v_{av}}=20\hat{i}+10\hat{j} m/s
In magnotude average velocity:
v_{av}=\sqrt{20^2+10^2}=10\sqrt{5}  m/s

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