Physics, asked by poudelarbin, 1 year ago

a ball is thrown vertically upwards covers equal distances in its 4th and 5th second. Then initial velocity of projection will be?

Answers

Answered by enoosap9
9

Answer:

40m/s

Explanation:

When a ball is thrown upwards if it occupies equal distance in successive time intervals then the ball is projects up in t second reaches the maximum height and in t+1 second it returns back. So,

distance travelled in 4th second (s4) = u-(g/2)(2t-1) = u-(10/2)*(2*4-1) = u-35

distance travelled in 5th second (s5) = u-(g/2)(2t-1) = u-(10/2)*(2*5-1) = u-45

as in the fifth second the body returns back so u is -ve and g is +ve.

So, s5= -u+45

According to the question,

s4 = s5

u-35 = -u+45

u=80/2 =40m/s

Answered by AnkitaSahni
0

Given :

Distance covered in 4th sec (S_{4th}) = Distance covered in 5th sec (S_{5th})

To Find :

Initial velocity of the projection (u)

Solution :

We know, The distance covered in nth second S_n_t_h can be given as,

     S_n_t_h = S_n - S_n_-_1

So, S_{4th} = S₄ - S₄_-_1       and  S₅_t_h = S₅ - S_5_-_1

Let 'u' be the initial velocity, 't' be the time taken 'g' be the acceleration due to gravity and S be the distance

Using Kinematic equation for uniformly accelerated motion, we have

      S_4_t_h = (ut₄ - \frac{1}{2}gt₄²)  -  (ut₃ -  \frac{1}{2}gt₃²)

or,   S_4_t_h = [u×4 -  \frac{1}{2}×g(4)²] - [u×3 -  \frac{1}{2}×g×(3)²]

or,   S_4_t_h = 4u - 8g - 3u + 4.5g

or,   S_4_t_h = u - 3.5g

Now,

      S_{5th}         = (ut₅ - \frac{1}{2}gt₅²) - S_4_t_h

or, 2S_4_t_h        = (u×5 - \frac{1}{2}×g×5²)              (As  S_4_t_h  =   S_{5th})

or, 2S_{4th}        = 5u - 12.5g

or, 2(u - 3.5g) = 5u - 12.5g

or, 2u - 7g      = 5u - 12.5g

or, 2u - 70      = 5u - 125

or,        3u       = 55

∴            u       = 18.33 m/sec

Therefore, the initial velocity is 18.33 m/sec.

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