A blue ball is thrown upward with an initial speed of 22 m/s, from a height of 0

Question

A blue ball is thrown upward with an initial speed of 22 m/s, from a height of 0.8 meters above the ground. 2.7 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9 m/s from a height of 27.3 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. 1) How long after the blue ball is thrown are the two balls in the air at the same height? 2) Which statement is true about the blue ball after it has reached its maximum height and is falling back down? a) The acceleration is positive and it is speeding up b) The acceleration is negative and it is speeding up c) The acceleration is positive and it is slowing down d) The acceleration is negative and it is slowing down

Explanation / Answer

1)

Applying r = ut + at^2/2

height of blue ball after time t will be

h = 0.8 + 22t - 9.81t^2/2

and height of red ball,

h = 27.3 - 9(t-2.7) - 9.81(t -2.7)^2/2

height is same for both ball then

0.8 + 22t - 9.81t^2/2 = 27.3 - 9(t-2.7) - 9.81(t -2.7)^2/2

0.8 + 22t - 9.81t^2/2 = 27.3 - 9t + 24.3 - 9.81t^2/2 + 26.487t - 35.757

4.513t = 15.043

t = 3.33 sec

2).
when blue ball is falling then velocity is in downward direction and acceleration is also in downward
direction.

when v and a both are in same direction then speed will increase.

hence it is speeding up.

if downward direction is taken as negative then a will be negative.

then ans will be (B) . .....Ans

(Otherwise ans will be (A) (if downward direction is taken as +ve))

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