A red ball is thrown down with an initial speed of 1 m/s from a height of 27 met

Question

A red ball is thrown down with an initial speed of 1 m/s from a height   of  27 meters above the ground.  Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward   with an initial speed of 24.3 m/s, from a height of 0.7 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 red ball is thrown are the two balls in the air at the same height?

6)Which statement is true regarding the blue ball?

Explanation / Answer

5)

y = -0.5 g t^2 + vo t + y0

red ball:

==> y1 = -0.5 * 9.8 * t^2 - 1 * t + 27

blue ball:

==> y2 = -0.5 * 9.8 * (t - 0.6)^2 + 24.3 * (t - 0.6) + 0.7

the two balls in the air at the same height ==> y1 = y2

==> -0.5 * 9.8 * t^2 - 1 * t + 27 = -0.5 * 9.8 * (t - 0.6)^2 + 24.3 * (t - 0.6) + 0.7

==> t = 1.37 s

6)

After it is released and before it hits the ground, the blue ball is sometimes moving faster than the red ball at any given time.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.