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Two balls of equal masses are thrown upward along the same vertical at an interval of 2 seconds with the same initial velocity of 40m/s.then these collide at a height of(take g=10m/s2)?
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Two balls of equal masses are thrown upward along the same vertical at...
Problem Statement

Two balls of equal masses are thrown upward along the same vertical at an interval of 2 seconds with the same initial velocity of 40m/s.then these collide at a height of(take g=10m/s2)? Explain in details.


Calculating the Time of Flight of the First Ball


  • Using the formula:
    h = ut + 0.5gt^2
    where h = height, u = initial velocity, g = acceleration due to gravity, and t = time

  • For the first ball, initial velocity (u) = 40m/s, acceleration due to gravity (g) = 10m/s^2, and height (h) = 0 (since it will return to the same height)

  • Substituting these values in the formula, we get:
    0 = (40)t - 0.5(10)t^2

  • Simplifying, we get:
    5t^2 - 40t = 0

  • Therefore,
    t = 0s or t = 8s (We discard t=0s as it is the time at which the ball is thrown)

  • Thus, the time of flight of the first ball is
    8 seconds



Calculating the Maximum Height of the First Ball


  • Using the formula:
    h = (u^2/2g)
    where h = maximum height, u = initial velocity, and g = acceleration due to gravity

  • Substituting the values, we get:
    h = (40^2/2*10)

  • Therefore,
    h = 80 meters



Calculating the Time of Flight of the Second Ball


  • Since the second ball is thrown after 2 seconds, its time of flight will be 2 seconds less than that of the first ball

  • Therefore, the time of flight of the second ball is
    6 seconds



Calculating the Maximum Height of the Second Ball


  • Using the same formula as before, we get:
    h = (40^2/2*10)

  • Therefore,
    h = 80 meters



Calculating the Time and Height of Collision


  • Since both balls have the same maximum height, they will collide at that height

  • Using the formula:
    h = ut + 0.5gt^2

  • For the first ball, substituting the values, we get:
    80 = (40)t - 0.
Community Answer
Two balls of equal masses are thrown upward along the same vertical at...
If we take mass a and mass b mass a takes t sec so mass b will take t-2 sec as it is dropped at an interval of 2 sec .so s for mass a will be .Taking sign convention position towards up position


S1= 40t-1/2*10t^2
= 40t - 5t^2


S2= 40(t-2)-1/2*10*(t-2)
= 5t^2-25t+20

if they will collide then S1=S2


on keeping them equal and solving them we will get t= 5 seconds

then substitute the value of t in any equation and the answer will be approx 73.5
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Two balls of equal masses are thrown upward along the same vertical at an interval of 2 seconds with the same initial velocity of 40m/s.then these collide at a height of(take g=10m/s2)?
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