Abaseball leaves a pitcher's hand horizontally at a speed of 104 km/h. the distance to the batter is 18.3 m. (ignore the effect of air resistance.) (a) how long does the ball take to travel the first half of that distance? (b) the second half? (c) how far does the ball fall freely during the first half? (d) during the second half?

a)t = 0.323 sec

b)t'= 0.323 sec

c)|h|= 0.52 m

d)y=1.56 m

Explanation:

The speed of ball ,U=104 km/h

U = 104 x 5/18  m/s =28.88 m/s

x= 18.3 m

a)

The half of distance = 18.3/2=9.15 m

We know that

Distance = Velocity x time

9.15 = 28.28 .t

t = 0.323 sec

b)

The total time to cover distance 18.3 m

18.3 = 28.88 x T

T=0.646 m

So the second half distance will cover in t' time

t' = T -t

t' = 0.646 - 0.323

t' = 0.323 s

c)

Initial velocity in y direction is zero.

Vo= 0

h= - 0.52 m         ( this is displacement and we know that it is a vector)

So magnitude of |h|= 0.52 m

d)

H= - 2.08 m

|H|= 2.08 m

So the in second half y = H- h

y= 2.08 - 0.52 m

y=1.56 m

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