two balls are projected simultaneously with the same speed from the top of a tower, one vertically upwards and the other vertically downwards. they reach the ground in 18 seconds and 8 seconds respectively. if the third ball is just released from the top of the tower , find the time required for it to reach the ground.
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sorry in eq 1 it is t1 at the place of t and in eq2 it is t2 at the place of t beacuse t is not same for both upward and downward motion...answer is after solving both the equations we get
t3 = sqrt of t1t2
t3 = square root of 18x8 i.e. of 144
hence t3 = 12 sec is the correct answer
-h = ut1 -1/2 gt1^2
-h = ut2-1/2 gt2^2
on multiplying eq1 with t2 and eq2 with t1 and after adding both the equations we get
-h (t1 +t2) = -1/2 g t1t2 (t1+t2)
h =1/2 g t1t2
and -h = -1/2 g t3^2
so t1t2 = t3 ^2
t3 = square root of t1 t2
sq root of ( 18 ×8)
sq root of 144
t3 = 12 sec correct answer
t3 = sqrt of t1t2
t3 = square root of 18x8 i.e. of 144
hence t3 = 12 sec is the correct answer
-h = ut1 -1/2 gt1^2
-h = ut2-1/2 gt2^2
on multiplying eq1 with t2 and eq2 with t1 and after adding both the equations we get
-h (t1 +t2) = -1/2 g t1t2 (t1+t2)
h =1/2 g t1t2
and -h = -1/2 g t3^2
so t1t2 = t3 ^2
t3 = square root of t1 t2
sq root of ( 18 ×8)
sq root of 144
t3 = 12 sec correct answer
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