a ball dropped foma first floor of a building reaches the ground in 0.9 seconds.taking g=10m/s2 what is the average speed of the ball during 0.9s?

Question

Naresh Kinha · Answer

Since we know that: V=u+at And it is given in the question that the body is dropped, so U=0 Now we know here a=10m/s and t=0.9sec V=10×0.9=9m/s

Supriyaa Hejib · Answer

u=0
t=0.9 sec
a=10 m/s^2
Therefore, using the second equation of motion,
s=ut+1/2a(t)^2
s=1/2*10*(0.9)^2
s=5*0.81
s=4.05m

Now, we know that average speed= Total distance/Total time
= 4.05/0.9
= 4.5 m/s

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a ball dropped foma first floor of a building reaches the ground in 0.9 seconds.taking g=10m/s2 what is the average speed of the ball during 0.9s?

a ball dropped foma first floor of a building reaches the ground in 0.9 seconds.taking g=10m/s2 what is the average speed of the ball during 0.9s?

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