A tennis ball is shot vertically upward inside a tower with an initial speed of 20.0 m/s. Neglect air resistance. Approximately how long does it take the tennis ball to reach its maximum height?
0.50 s
2.04 s
4.08 s
6.08 s
9.80 s

Determine the velocity of the ball 3.0 s after it is thrown
9.40 m/s, downward
9.40 m/s, upward
29.4 m/s, downward
38.8 m/s, upward
38.8 m/s, downward

The ball's height at time t is

y = (20.0 m/s) t - 1/2 g t²

where g is the acceleration due to gravity, with magnitude 9.80 m/s².

Also, recall that

v² - u² = 2 a ∆y

where u is the initial velocity, v is the final velocity, a is the acceleration, and ∆y is the change in height. Let Y be the maximum height. At this height, v = 0, so

- (20.0 m/s)² = 2 (-g) Y

==>  Y ≈ 20.408 m

Plug this into the first equation and solve for t :

Y = (20.0 m/s) t - 1/2 (9.80 m/s²) t²

==>  t ≈ 2.04 s

The ball's velocity at time t is

v = 20.0 m/s - g t

After t = 3.0 s, its velocity will be

v = 20.0 m/s - (9.80 m/s²) (3.0 s)

v = -9.40 m/s

or 9.40 m/s in the downward direction.

the answer is b

explanation: