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Grade: 12th pass
        
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 
7 months ago

Answers : (2)

Sahani Kumar
99 Points
							
Total time period (Ť) = 2u/g 
Maximum height H =( u^2)/(2g) =(( T^2)*g)/2
Now,
initial velocity u= Tg/2  
Final velocity (v) =?  acceleration (a) = g
Time =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)/4
Now 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) /2
So h= 5H/16.
Regards 
Askiitians member
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7 months ago
Atul Nagar
34 Points
							
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=0
Using first equation f motion
v=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/4
H(T/4)=u(T/4)-1/2(g)(T^2/16)
since u=gT/2
H(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 1
If we want to represent this height in terms of H, using III eqn for motion
u^2=2gH =>  ( g^2.T^2)/4=2gH
T^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 
 
 
 
 
7 months ago
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