Question

A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upward from the ground with a velocity of 25m/s . Calculatate when and where the two stone will meet .​

Answer 1

Explanation:

100-S) = distance travelled by the projected stone. Therefore, Two stones will meet after 4 s. = 80 m. Thus , both the stone will meet at a distance of 80 m from the top of tower

Answer 2

$here$

$height \: of \: the \: tower \: is \: 100 \: m \: . \:$

$now \: suppose \: the \: two \: stones \: meet \: at \: a \: point \: p \:which \: is \: at \: a \: height \: x \: above \: the \: ground \: . \:$

$so \: that \: the \: distance \: of \: p \: from \: the \: top \: of \: the \: tower \: is \: 100 - x$

$i) \: for \: the \: stone \: falling \: from \: top \: of \: tower =$

$height \: (h) = (100 - x) \: m$

$initial \: velocity(u) = 0$

Time (t)=?

$and$

$acceleration \: due \: to \: gravity(g) = 9.8 \: m {s}^{ - 2}$

$now$

$h = ut + \frac{1}{2} g {t}^{2}$

$so$

$100 - x = 0 \times t + \frac{1}{2} \times 9.8 \times {t}^{2}$

$or$

$100 - x = 4.9 {t}^{2}$

$ii) \: for \: stone \: projected \: vertically \: upward \: =$

$height(h) = x \: m$

$initial \: velocity(u) = 25 \: m {s}^{ - 1}$

$and$

$acceleration \: due \: to \: gravity(g) = - 9.8 \: m {s}^{ - 2}$

$now$

$st = ut + \frac{1}{2} g {t}^{2}$

$so$

$x = 25 \times t + \frac{1}{2} \times ( - 9.8) \times {t}^{2}$

$x = 25t - 4.9 {t}^{2}$

$on \: adding \: equation \: (1) \: and \: (2) \: we \: get =$

$100 - x + x = 4.9 {t}^{2} - 4.9 {t}^{2}$

$or$

$100 = 25 \: t$

$t = \frac{100}{25}$

$t = 4 \: s$

Thus, the two stones will meet after a time of 4 seconds.

Now, from equation (1) we have :

$100 - x = 4.9 {t}^{2}$

Putting t=4 in equation, we get :

$100 - x = 4.9 \times {(4)}^{2}$

$100 - x = 4.9 \times 16$

$100 - x = 78.4$

$100 - 78.4 = x$

$21.6m = x$

or,

$x = 21.6$

Thus , the two stones will meet at a height of 21.6 meters from a ground .