Question

A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 2.90 seconds for the ball to reach its maximum height. a) Find the balls initial velocity b) Find the height it reaches

Answer 1

The initial velocity of the baseball is 28.449 m/s, and the maximum height it reaches is approximately 41.3 meters.

Step-by-step explanation:

Finding the Initial Velocity and Maximum Height of a Baseball

To find the initial velocity (vi) of the baseball, we use the following kinematic equation for uniformly accelerated motion, which in this case is due to gravity (g = 9.81 m/s2):

v = vi - gt

At the maximum height, the final velocity (v) is 0 m/s, and the time (t) taken to reach there is 2.90 seconds, as observed by the fan. Plugging in these values gives us:

0 = vi - (9.81 m/s2)(2.90 s)

vi = (9.81 m/s2)(2.90 s)

vi = 28.449 m/s

To find the maximum height (h) the ball reaches, we use the following equation:

h = vit - ½gt2

By substituting the known values:

h = (28.449 m/s)(2.90 s) - ½(9.81 m/s2)(2.90 s)2

h = 41.3015 m - 41.29845 m

h = 41.3 m (rounded to one decimal place)

Answer 2

We'll assume that the whole scene takes place on Earth.
So the acceleration of gravity is 9.8 m/s² downward.

a). Gravity makes anything fall 9.8 m/s faster every second.
That's the same thing as rising 9.8 m/s slower every second.

If it takes 2.9 seconds to reach its maximum height, then
it must have started out rising at (9.8 x 2.9) = 28.4 m/s.

b). The ball left the bat at 28.4 m/s.
After 2.9 seconds, its speed was zero. (That's why it started falling.)
It's average speed during the climb was

1/2 (28.4 + 0) = 14.2 m/s .

It rose straight up at an average speed of 14.2 m/s for 2.9 seconds,
so it reached a maximum height of

(14.2 x 2.9) = 41.2 meters .