A blue ball is thrown upward with an initial speed of 20.2 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 20.2 m/s, from a height of 0.8 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.6 m/s from a height of 22.8 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

5) How long after the blue ball is thrown are the two balls in the air at the same height?

6) Which plot correctly shows the velocity of the two balls as a function of time?

7) Which statement is true about the blue ball after it has reached its maximum height and is falling back down?

Explanation / Answer

Solution:-

let the time be t

let x be the distance moved by blue ball
x=20.2*t-0.5*9.81*t^2..........eq1
20-x=6.6*(t-2.5)+0.5*9.81*(t-2.5)^2............eq2

solving eq1 and eq2
20=26.8*t-16.5+30.65-24.52*t
2.28*t=5.85
t=2.56 sec

6)B shows the velocity of two balls as a function of time
7)The acceleration is positive and it is speeding up

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