please draw the graph and put all the information in the graph then solve and sh

Question

please draw the graph and put all the information in the graph then solve and show all the steps

1A baseball is bunted down tha third-basa ina leaing tha cat feom a height of 0.gcom with an initial velocity o 3.00mi's at an argis o gac abave the harizcntal. (a) How far does the baseball travel before hiting te ground? (b) What is the speed of the ball just befora t hits the ground? (c) What is the maximum height of tha baseball above ta hafzontal in this traval ? (d) What is the ball normal (centripetal) accaleration atthe maimum hight2 (e) What is the baseball tangential acceleration at the maimum height 2 Answer the questions (d) and (e) for tha instart just betora te ba hits the (g) What are the and at the paint of the grounding ? (a) What are the tadiuges of the baseball path currature at helgt maimum

Explanation / Answer

h = height from which ball is dropped = 0.9 m

initial velocity = Vo = 3 m/s

angle = theta = 30

Along the y-direction ::

initial velocity = Voy = Vo Sin30 = 3 Sin30 = 1.5 m/s

acceleration = ay = -9.8 m/s2

h = height from which ball is dropped = -0.9 m

Using the equation ::

h = Voy t + (0.5) ay t2

-0.9 = 1.5 t + (0.5) (-9.8) t2

t = 0.61 sec

along the X-direction ::

initial velocity = Vox = Vo Cos30 = 3 Cos30 = 2.6 m/s

acceleration = ax = 0 m/s2

time = t = 0.61 s

Using the equation :

X = Vox t + (0.5) ax t2

X = (2.6) (0.61) + (0.5) (0) t2

X = 1.586 m

b)

along the Y-diection

Vfy= Voy + ayt

Vfy = 1.5 + (-9.8) (0.61)

Vfy = - 4.48 m/s

Vfx = 2.6 m/s

net velocity :

V = sqrt (Vfy2 + Vfx2)

V = sqrt ((-4.48)2 + (2.6)2)

V = 5.18 m/s

c)

Using the eqation:

V2fy= V2oy + 2ayH

02 = 1.52 + 2 (-9.8)H

H = 0.115 m

d)

a = 9.8 m/s2

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