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A body is thrown vertically upwards from the top of a tower from point A. It reaches the ground in time t1. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time t 2, if it is allowed to fall freely from A, then the time it takes to reach the ground is given by

A body is thrown vertically upwards from the top of a tower from point A. It reaches the ground in time t1. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time t 2, if it is allowed to fall freely from A, then the time it takes to reach the ground is given by

Grade:9

4 Answers

Khimraj
3007 Points
7 years ago
Write three equation and solve these equation 1. h = -ut1 + 1/2(gt1²). 2. h = ut2 +1/2(gt2²). 3. h = 1/2gt². Now solve these equation then u get t =√(t1t2). I give suggestions to solve equations multiply first by t2 and second by t1 and add them.Hope it clears. If still u have doubts then please clarify. If u like answer then please approve the answer.
Ramakant Dogiwal
19 Points
6 years ago
h=ut1−12gt21−h=ut1−12gt12--------(i)h=ut2+12h=ut2+12gt22gt22--------(ii)Where h is a positive number. Multiplying (i) by t2t2 and (ii) by t1t1 we get,h(t1+t2)=12=12h(t1+t2)=12=12gt1t2(t1+t2)gt1t2(t1+t2)-----------(iii)y=12y=12gt2gt2---------(iv)From (iii) and (iv)t=t1t2−−−√t=t1t2Hence D is the correct answer.
Nandini
13 Points
5 years ago
Suppose the body was thrown with a velocity vand in t' time it reached at the top (of height h' from a) and then starts to fall again and in another t'' time it touches the ground. We know that when it will reach at top, the velocity will be zero, so we have
0=vgtt=vg
Again we have
02=v22gh'h'=v22g
Now while coming down total height will be H=h+h', so we have
h+h'=12gt''2h+v22g=12gt''2h+12gv2g2=12gt''2h+12gt'2=12gt''2h=12g(t''2t'2)=12g(t''t')(t''+t')=12gt1(t''t')
Now when the body again come back to point a, it will again have same velocity and it will come to that point a in t' time only. Also when it will come back to a with velocity v and starts falling, it is equivalent to throwing a ball with velocity v in downwards direction, so time taken will be equal to t, we have t2=(t''-t') Hence the above equation becomes
h=12gt1t2
Now using the free fall equation, supposing free fall from a will take t time, we have
h=12gt2
So we have
12gt2=12gt1t2t=t1t2
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

645-2393_0101.pngFrom fig(A):
−h = ut1​ − 0.5​g(t1)^2
​⇒ h/t1 ​= −u +0.5​gt1​..........(i)

From fig(B):
−h = −ut2 ​− 0.5​g(t2)^2​
⇒ ​h/(t2) ​= u + 0.5​gt2​.........(ii)

From fig(C):
−h = −0.5​g(t3)^2​
⇒ h = 0.5​g(t3)^2​.........(iii)

From (i) and (ii):
h[1/(t1) ​+ ​1/(t2)​] = 0.5​g(t1​+t2​)........(iv)

∴ By eq.(iii) and (iv)
0.5g(t3)^2 ​= 0.5​gt1​t2​
⇒ t3 ​= square root (t1​t2)​​

Thanks and Regards

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