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Question

# A cricket ball thrown across a field is at heights h1 and h2 from the point of projection at times t1 and t2 respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is:

A
h1t22h2t21h1t2h2t1
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B
h1t21+h2t22h2t1+h1t2
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C
h1t22+h2t21h1t2+h2t1
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D
h1t21h2t22h1t1h2t2
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Solution

## The correct option is A h1t22−h2t21h1t2−h2t1For vertical movement of the ball h1=usinθt1−12gt21or t1=h1+12gt21usinθ .....(i)Similarly, h2=usinθt2−12gt22 or t2=h2+12gt22usinθ ......(ii)Divide equation (i) by equation (ii) t1t2=h1+12gt21usinθh2+12gt22usinθ⇒h1t2−h2t1=g2(t1t22−t21t2)The time of flight of the ball T=2usinθg=2g(usinθ) [from equation (i)] =2g⎡⎢ ⎢ ⎢⎣h1+12gt21t1⎤⎥ ⎥ ⎥⎦=2t1[h1g+t212] =h1t1×2g+t1=h1t1×(t1t22−t21t2h1t2−h2t1)+t1 =h1t1t22−h1t21t2+h1t21t2−h2t31t1(h1t2−h2t1) =(h1t22−h2t21h1t2−h2t1)

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