You visited us 11 times! Enjoying our articles? Unlock Full Access!

Motion Under Gravity

From the top ...

Question

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t_{3} = \sqrt{t_{1} t_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{t_{3}} = \frac{1}{t_{2}} - \frac{1}{t_{1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $t_{3} = \sqrt{t_{1} t_{2}}$
Applying equations of motion for all three cases
$- h = u t_{1} - \frac{1}{2} g t_{1}^{2} . . . (1)$
$- h = - u t_{2} - \frac{1}{2} g t_{2}^{2} . . . (2)$
$& - h = - \frac{1}{2} g t_{3}^{2} . . . (3)$
Multiply equation 1 with $t_{2}$ and equation 2 with $t_{1}$ and add them
We get,
$\Rightarrow h = \frac{1}{2} g t_{1} t_{2} . . . . . . . (4)$
Now by (3) & (4)
$t_{3}^{2} = t_{1} t_{2}$
$t_{3} = \sqrt{t_{1} t_{2}}$

Suggest Corrections

Similar questions

t_{1}

t_{2}

t_{3}

5

s

1

s

u

H

t_{1}

t_{2}

t_{3}

t_{3}

t_{1}

t_{2}

u

t_{1}

t_{2}

θ

t_{1}

θ

t_{2}

t_{3}

Join BYJU'S Learning Program