Question

Please answer this fast. No spams please.

Answer 1

Answer:

b) time=2 sec and speed=20 m/s

Explanation:

DEAR STUDENT,

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b) GIVEN: h = 20m

                 u = 0m/s

   TO FIND: t = ?

                    v = ?

   SOLUTION:

   FORMULAE APPLIED:  1) h = ut + 1/2gt^2

                                           2) v = u + gt

      h = ut + 1/2gt^2

=> 20 = 1/2 x 10t^2

=> 40 = 10t^2

=>   4 = t^2

=>   t  = 2 sec       .......1

Now,

      v = u + gt

      v =  10 x 2        .......(FROM 1)

      v = 20 m/s

CONCLUSION: The time is 2 seconds and speed is 20 m/s

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REGARDS.

Answer 2

DEAR VAMP.

SOLUTION

FIRSTLY I MAKE SURE THAT INITIAL VELOCITY OF THE BIKE IS ZERO .

let t be the time in which the theif was caught by the policemen

distance travelled by the theif

s=ut+1/2at^2. (u=0)

s=1/2at^2.....(1)

distance travelled by policemen to catch theif

s+d=ut+1/2at^2 (a=0)

s + d = ut......(2)

where u is the velocity of the policemen

put (1) in (2)

d + 1/2at^2 = ut

2d + at^2 -2ut = 0

at^2 - 2ut + 2d = 0

now you find a quadratic equation whose root is always real because time can never be imaginary.

so D>0

4u^2 - 8ad > 0

u^2 > 2ad

(B)

As particle dropped from height so initial velocity of the particle is zero.

acc to third equation of motion.

v^2 = u^2 + 2gh

v^2 = 0 +2(10)(20)

v^2 = 400

v = 20m/sec

acc to first equation of motion.

v = u + gt

20 = 0 + (10)t

20 = 10 t

time = 2sec