Question

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?

Answer 1

Answer:

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.