At a hot air balloon race, a person on the ground shots a ball of confetti from a cannon to start the race. One of the balloons is rising with a constant speed of 11m/s. The confetti ball is shot completely verically with an initial speed of 18m/s. a) how much time will it take for the ball and balloon to meet? b) where, above the ground, will the ball reach the balloon?

a) The balloon and the ball will meet after 1.43 s.

b) The ball will reach the balloon at 15.7 m above the ground.

Explanation:

The height of the confetti ball is given by the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The height of the ball is given by this equation:

y = y0 + v · t

Where v is the constant velocity.

When the ball and the ballon meet, both heights are equal. Let's consider the ground as the origin of the frame of reference so that y0 = 0:

y balloon = y ball

y0 + v · t = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

11 m/s · t = 18 m/s · t - 1/2 · 9.8 m/s² · t²

0 = - 4.9 m/s² · t² + 18 m/s · t - 11 m/s · t

0 = - 4.9 m/s² · t² + 7 m/s · t

0 = t (-4.9 m/s² · t + 7 m/s)

t = 0 and

0 = - 4.9 m/s² · t + 7 m/s

-7 m/s / - 4.9 m/s² = t

t = 1.43 s

They will meet after 1.43 s

b) Now let's calculate the height of the balloon after 1.43 s

y = v · t

y = 11 m/s · 1.43 s = 15.7 m

The ball will reach the balloon at 15.7 m above the ground.