a ball is thrown vertically upwards covers equal distances in its 4th and 5th second. Then initial velocity of projection will be?
Answers
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
Given :
Distance covered in 4th sec () = Distance covered in 5th sec ()
To Find :
Initial velocity of the projection (u)
Solution :
We know, The distance covered in nth second S can be given as,
S = S - S
So, S = S₄ - S₄ and S₅ = S₅ - S
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 = (ut₄ - gt₄²) - (ut₃ - gt₃²)
or, S = [u×4 - ×g(4)²] - [u×3 - ×g×(3)²]
or, S = 4u - 8g - 3u + 4.5g
or, S = u - 3.5g
Now,
S = (ut₅ - gt₅²) - S
or, 2S = (u×5 - ×g×5²) (As S = S)
or, 2S = 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.