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

(a) 29.0m/s

(b) 42.05m

Explanation:

(a) Since the motion is in one dimension (i.e vertical), we can use one of the equations of motion as follows;

v = u + at            ------------(i)

Where;

v = final velocity of the baseball at a time t

u = initial velocity of the baseball at launch time

a = acceleration due to gravity

t = time taken to reach a certain height

Now;

At maximum height;

v = 0  (velocity is zero when the baseball reaches maximum height)

t = 2.90s

a = -g = -10m/s² (negative sign because the base ball moves upwards against gravity)

Substitute these values into equation (i) as follows;

0 = u - 10(2.90)

u = 29.0m/s

Therefore, the initial velocity of the baseball is 29.0m/s

(b) To get the height reached we use another equation of motion as follows;

Using one of the equations of motion as follows;

v² = u² + 2as             -----------------(ii)

Where;

v = final velocity of the baseball

u = initial velocity of the baseball

a = acceleration of the baseball

s = vertical distance covered by the baseball

Remember that;

At maximum height,

v = 0

Also,

u = 29.0m/s     (as calculated above)

a = -10m/s²

Substitute these values into equation(ii) as follows;

0² = 29.0² + 2(-10)s

0 = 841 - 20s

841 = 20s

s =  

s = 42.05m

Therefore, the maximum height reached is 42.05m

1/0.14nm = 10^10/1.4 a lot