Question

Only the brainliest ones can solve this jabardast question

Answer 1

Answer:

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Answer 2

Answer:

Given ,

u= 10 m/s, a= g= -10 m/s²,s= - 40 m

By second equation of motion,

S= ut+1/2at²

Or-40= 10*t+ 1/2*-10*t²

Or, 5t²-10t-40= 0

Or,t²-2t-8= 0

Or, t²-4t+2t-8=0

Or,t(t-4)+2(t-4)=0

Or,(t-4)(t+2)=0

By zero product rule,

t-4=0. t-2=0

Or, t=4 Or,t=-2.

Since time cannot be negative, t= 4 seconds.

While returning from the highest point , u= 0, total height travelled= 5+40= 45m.

By the third equation of motion,

V²=u²+2gh

Or, v= √ 0+2*10*45

Or, v= √ 900

Or, v= 30 m/s

Distance covered by ball in the course of the journey:

V²=u²-2gh

Or, h=u²/2g( at highest point,v=0)

Or, h=100/2*10

Or, h= 5 m

Hence , total distance= 2*5 = 10 m

Hence, average speed= total distance/total time = 10/4=2.5 m/s.

Total Displacement of the ball= 10-40=-30 m

Hence, average velocity= total Displacement/ total time= -30/4= - 7.5 m/s = 7.5 m/s in opposite direction.