From the top of tower, a stone is thrown up. It reaches the ground in t1 s. A second stone thrown down with the same speed reaches the ground in t2 s. A third stone released from rest reaches the ground in t3 s. Then

Question

From the top of tower, a stone is thrown up. It reaches the ground in t1 s. A second stone thrown down with the same speed reaches the ground in t2 s. A third stone released from rest reaches the ground in t3 s. Then (A) t3 = ((t1+t2)/2) (B) t3 = √t1t2 (C) (1/t3) = (1/t1) + (1/t2) (D) t23 = t22 - t21

Tardigrade Admin · Accepted Answer

When stone is thrown vertically upwards from the top of tower of height h, then

h = u t_{1} + \frac{1}{2} g t_{1}^{2} ... (i)

When stone is thrown vertically downwards from the top of tower, then

h = u t_{2} + \frac{1}{2} g t_{2}^{2} ... (ii)

When stone is released from the top of tower, then

h = \frac{1}{2} g t_{3}^{2} ... (iii)

From equation

(i)

, we get

\frac{h}{t _{1}} = - u + \frac{1}{2} g t_{1} ... (i v)

From equation

(ii)

, we get

\frac{h}{t _{2}} = - u + \frac{1}{2} g t_{2} ... (v)

Adding equations

(i v)

and

(v)

. we get

h (\frac{1}{t _{1}} + \frac{1}{t _{2}}) = \frac{1}{2} g (t_{1} + t_{2})

or

h = \frac{1}{2} g t_{1} t_{2}

Putting the value in equation

(iii)

, we get

t_{3} = t_{1} t_{2}

Q. From the top of tower, a stone is thrown up. It reaches the ground in $t_{1}$ s. A second stone thrown down with the same speed reaches the ground in $t_{2}$ s. A third stone released from rest reaches the ground in $t_{3}$ s. Then

$t_{3} = \frac{( t _{1} + t _{2} )}{2}$

$t_{3} = t_{1} t_{2}$

$\frac{1}{t _{3}} = \frac{1}{t _{1}} + \frac{1}{t _{2}}$

$t_{3}^{2} = t_{2}^{2} - t_{1}^{2}$

Q. From the top of tower, a stone is thrown up. It reaches the ground in t1​ s. A second stone thrown down with the same speed reaches the ground in t2​ s. A third stone released from rest reaches the ground in t3​ s. Then

t3​=2(t1​+t2​)​

t3​=t1​t2​​

t3​1​=t1​1​+t2​1​

t32​=t22​−t12​

Q. From the top of tower, a stone is thrown up. It reaches the ground in $t_{1}$ s. A second stone thrown down with the same speed reaches the ground in $t_{2}$ s. A third stone released from rest reaches the ground in $t_{3}$ s. Then

$t_{3} = \frac{( t _{1} + t _{2} )}{2}$

$t_{3} = t_{1} t_{2}$

$\frac{1}{t _{3}} = \frac{1}{t _{1}} + \frac{1}{t _{2}}$

$t_{3}^{2} = t_{2}^{2} - t_{1}^{2}$