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`        A child tosses a ball directly upward. Its total time in the air is T. Its maximum height is H. What is its height after it has been in the air a time T/4? Neglect air resistance. PLZ clear this tricky question `
one year ago

```							Total time period (Ť) = 2u/g Maximum height H =( u^2)/(2g) =(( T^2)*g)/2Now,initial velocity u= Tg/2  Final velocity (v) =?  acceleration (a) = gTime =T/4           Height (h) is to be found=?Now from equation v= u + a t  we can find final velocity. v= (Tg/2)+(Tg/4) = (3Tg)/4Now from equation s= (u)*(t )+ 1/2(a*t^2) we can find (h) So h= (Tg/2)*(T/4) + 1/2((g*T^2)/16)      h=((T^2)*g)/(8) + ((T^2)*g)/(32)      h=((5T^2) * g)/(2 * 16)   we know that H=((T^2 )* g) /2So h= 5H/16.Regards Askiitians memberIf you like this answer please encourage us by approving this answer.
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one year ago
```							We know that Time of Ascent = Time to Descent, That means it will take time T/2 to reach maximum height H.This indicates that the object will be in Upward journey at t=T/4.Let initial velocity is u, a= – g and at H , v=0Using first equation f motionv=u-gT/2=0 This means u=gT/2--------(i)That meas the object is projected with speed of gT/2 in upward direction.Distance travelled in time t=T/4H(T/4)=u(T/4)-1/2(g)(T^2/16)since u=gT/2H(T/4)=gT^2/8-gT^2/32  H(T/4)= (3gT^2)/32  We can represent this height in terms of given variables H &T --------Ans 1If we want to represent this height in terms of H, using III eqn for motionu^2=2gH =>  ( g^2.T^2)/4=2gHT^2=8xH/g ---------------------------------(ii)Putting in Ans 1 we have H(T/4)=(3g/32)x(8H/g)We have H(T/4)=(3/4)H= 0.75 H Askiitian member B Tech
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one year ago
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