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

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach

at right angles to each other