Question

1.5x 10 m
17. A stone is allowed to fall from the top of a tower 100 m high
and at the same time another stone is projected vertically
upwards from the ground with a velocity of 25 m/s. Calculate
when and where the two stones will meet.​

Answer 1

Explanation:

i think that this relation can help u

Answer 2

Let the two stones meet after a time $t.$

$\bf (i).For\: the\: stone \:dropped\: from \:the\: tower:$

$\bold{\underline {Given:}}$

✯Initial velocity,$u = 0$

✯Time, $t ⟹=\:?$

✯Acceleration due to gravity, $g = 9.8 \:m\: {s}^{2}$

Let the displacement of the stone in time $t$from the top of the tower be $\:s$.

$⟹ s = ut+\dfrac{1}{2}gt^2$

$⟹0 ×t+\dfrac{1}{2}×9.8×t^2$

$∴\:\:\:\:\:\:\:⟹s\:=\:4.9t^2$.............(1)

$\bf(ii).For\: the\: stone \:thrown\: upwards :$

Initial velocity,$u = 25\: m\: {s}^{-1}$

Let the displacement of the stone from the ground in time $t$be$s'$.

Acceleration due to gravity, $g = - 9.8\: m\:{s} ^{-2}.$

From the equation of motion,

$⟹ s = ut+\dfrac{1}{2}gt^2$

$⟹ s' = ut+\dfrac{1}{2}gt^2$

$⟹ 25t+\dfrac{1}{2}×(-9.8)×t^2$

$∴\:\:\:\:\:⟹ s' = 25t-4.9\:t^2$..........(2)

The combined displacement of both the stones at the meeting point is equal to the height of the tower 200 m.

$∴\:\:\:\:\:\:\:\:\:⟹s+s'=100$

$⟹ \dfrac{1}{2}gt^2+25t-\dfrac{1}{2}gt^2=100$

$∴\:\:\:\:\:\:\:\:\:⟹t=\dfrac{100}{25}=4\:s$

In 4 s, the falling stone has covered a distant given by equation (1)as

$⟹s=4.9×4^2$

$=78.4\:m$

Therefore, the stones will meer after 4 s at a height (100 - 78.4)=21.6 m from the ground.