Question

A stone is released from the top of tower 100m. high and at the same instant another stone is throw n up from the base of the tower with a speed of 25m/s find when and where they meet in air

Answer 1

Heya user !!

Here's the answer you are looking for

(See the attachment for reference)

Let them eat at sea that is it and height of h from the ground after time t.

So, in time t the distance covered by the first stone is (100 - h) and the distance covered by the second stone is h.

☛For the 1st stone (at the top of the tower)

initial velocity (u) = 0
time = t
distance (s) = 100 - h
acceleration (a) = g

$Since, \: s = ut + \frac{1}{2} a {t}^{2} \\ \\ 100 - h = 0 + \frac{1}{2} g {t}^{2} \\ \\ h = 100 - \frac{1}{2} g {t}^{2}$

☛ For the second stone (at the bottom of the tower)

initial velocity (u) = 25m/s
time = t
distance (s) = h
acceleration (a) = (-g)

$h = 25t + \frac{1}{2} ( - g) {t}^{2} \\ \\ h = 25t - \frac{1}{2}g {t}^{2}$

From both the equations of h, we get,

$100 - \frac{1}{2} g {t}^{2} = 25t - \frac{1}{2} g {t}^{2} \\ \\ 100 = 25t \\ \\ t = \frac{100}{25} = 4$

So, after 4sec both the stones will meet.

Now put the value of t in any of the equations of h to get the distance.

$h = 100 - \frac{1}{2} g {t}^{2} \\ \\ h = 100 - \frac{1}{2} \times 10 \times 16 \\ \\ h = 100 - 10 \times 8 = 100 - 80 = 20$

➡️Therefore, the stones will meet at the height of 20m above the ground after 4sec.


★★ HOPE THAT HELPS ☺️ ★★