A blue ball is thrown upward with an initial speed of 19.6 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 19.6 m/s, from a height of 0.5 meters above the ground. 2.4 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 8 m/s from a height of 22.1 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/s? 1) What is the speed of the blue ball when it reaches its maximum height? m/s Submit 2) How long does it take the blue ball to reach its maximum height? s Submit 3) What is the maximum height the blue ball reaches? m Submit 4) What is the height of the red ball 3.12 seconds after the blue ball is thrown? m Submit 5) How long after the blue ball is thrown are the two balls in the air at the same height? s Submit

Explanation / Answer

1)

as the blue ball travels upward , it slows down due to acceleration in opposite downward direction. at the highest point, the ball comes to a momentary stop and then returns.

hence the velocity at the maximum height is zero.

2)

t = time taken by blue ball to reach maximum height

Using the equation

vfb = vib + at

0 = 19.6 + (-9.8) t

t = 2 sec

3)

for blue ball :

vib = initial velocity of blue ball = 19.6 m/s

Xib = initial position of blue ball = 0.5 m

vfb = final velocity at the maximum height = 0 m/s

Xfb = final position at maximum height

a = acceleration = - 9.8 m/s2

Using the equation

vfb2 = vib2 + 2 a (Xfb - Xib )

02 = 19.62 + 2 (- 9.8) (Xfb - 0.5 )

Xfb = 20.1 m

4)

t = time of travel for red ball = 3.12 - 2.4 = 0.72 sec

vir = initial velocity of red ball = - 8 m/s

Xir = initial position of red ball = 22.1 m

Xfr = final position at time "t"

a = acceleration = - 9.8 m/s2

Using the equation

Xfr = Xir + vir t + (0.5) a t2

Xfr = 22.1 + (- 8) (0.72) + (0.5) (- 9.8) (0.72)2

Xfr = 13.8 m

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