Question

plz answer this question ​

Answer 1

Answer:

Answer:

Given : 

Case 1:

For the ball A :

initialSpeed=u=20m/s[upwards]=-20m/s

a=10m/s²[downwards]=+10m/s²

time=t=tsec

Distance=Sa meters

From seond equation of motion,

Sa=ut+1/2at²

Sa=-20t+1/2x10xt²

Sa=-20t+5t² --------------equation(1)

Case II:

For ball B:

Initial speed=u=20m/s[downwards]=+20m/s

a=10m/s²[downwards]=+10m/s²

time =t sec

Distance =Sb meters.=(40-sa)

From seond equation of motion,

Sb=ut+1/2at²(4-Sa)

=20t+5t² ====[equation2]

Therefore , from equation 1 and 2 we get:

40=-20t+5t²+20t+5t²40

=10t²4=t²

t=2sec

Let us find out distance Sa in 2 sec

Sa=ut+1/2at²

sa=-40+20=-20m

From the ground that is at 20m the two balls will collide.

Answer 2

$\sf \Large Solution!!$

$\sf \large Given:$

$\sf \to Time\:taken\:to\:reach\:the\:ground\:by\:ball\:X,\:t_{x}=6\:seconds$

$\sf \to Time\:taken\:to\:reach\:the\:ground\:by\:ball\:Y,\:t_{x}=2\:seconds$

$\sf \to g=10\:m/s^{2}$

$\sf \large To\:find:$

$\sf \to Height\:of\:the\:tower(H)$

$\sf \to Initial\:speed\:of\:each\:ball$

$\sf \large Formula:$

$\sf \boxed{h=ut+\frac{1}{2}at^{2}}$

$\sf \large So,$

$\sf \to -H=v\times 6-\frac{1}{2}g\times 6^{2}...(i)$

$\sf \to -H=-v\times 2-\frac{1}{2}g\times 2^{2}...(ii)$

$\sf \large On\:subtracting\:(i)\:and\:(ii),\:we\:get,$

$\sf \to 0=8v-\frac{1g}{2}\times 36+\frac{1}{2}\times g\times 4$

$\sf \to V=\frac{18g-2g}{8}=\frac{\cancel{16}g}{\cancel 8}=2g=2\times 10= 20\:m/s^{2}$

$\sf \large Putting\:the\:value\:of\:V\:in\:(ii),\:we\:get,$

$\sf \to \cancel -H=\cancel -20\times 2\cancel -\frac{1}{2}\times 10\times 4$

$\sf \to H=40+20=60\:m$

$\sf \large Answers:$

$\sf \bigstar Height\:of\:tower=60\:m$

$\sf \bigstar Initial\:speed\:of\:each\:ball=20\:m/s^{2}$

$\sf \blue{Have\:a\:good\:day:)}$