Question

please solve the question.... don't SPAM​

Answer 1

Refer to the attachment for the figure.

Solution :

Assumptions-

A = man on Tower top

AB = height of Tower

E = car's first position

D = car's second position

Time taken to cover CD = 6 sec

To find - Time for crossing BC.

Speed is uniform, let's assume it x km/hr.

distance

speed = -------------

Time

=> d = speed × t

$= > ED= x \times 6 = 6x \\ bd = x \times t \\ \\ in \: ABC \\ \tan(60) = \frac{AB}{BD} \\ = > \sqrt{3} = \frac{h}{xt} \\ = > h = \sqrt{3} xt \\ \\ similarly \: in \: ABD \\ \tan(30) = \frac{AB}{ED + BD} \\ = > \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} xt}{6x + xt} \\ = > 6x + xt = \sqrt{3} ( \sqrt{3} xt) \\ = > 6x + xt = 3xt \\ = > 3xt - 2xt = 6x \\ = > \frac{2xt}{x} = 6 \\ = > 2t = 6 \\ = > t = \frac{6}{2} = 3$

Thus, time taken by the car to reach the foot of the Tower is 3 seconds.

Answer 2

Answer:

3 seconds

lets:

Speed is V

Time taken from 60 deg to 90 degree is T

Tower hieght is H

Cot30 - Cot 60 = V(t+6)/H - Vt/H= 6V/H

V/H = (root 3)/9

t= cot 60 / ( V/H) = 3 Second