From the window of a building, a bal is tossed from a height y_0 above the groun

Question

From the window of a building, a bal is tossed from a height y_0 above the ground with an initial velocity of 8.10 m/s and angle of 19.0 degree below the horizontal. It strikes the ground 5.00 s later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (Use the following as necessary: y_0. Assume SI units. Do not substitute numerical values; use variables only.) (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. (c) Find the equations for the x- and y-components of the position as functions of time. (Use the following as necessary: y_0 and t. Assume SI units.) (d) How far horizontally from the base of the building does the ball strike the ground? (e) Find the height from which the ball was thrown. (f) How long does it take the ball to reach a point 10.0 m below the level of launching?

Explanation / Answer

initial velocity of the ball vo= 8.1 m/s

angle with below the horizontal = 190

time t = 5 s

a)

initial coordinates of the ball is,

x-coordinate : x = 0 m

y -coordinate : y = h m

coordinate : (x,y) = (0 m , h m)

b)

x-component of the initial velocity of the ball is,

        v0x = v0 cos190

y-component of the initial velocity of the ball is

        v0y = v0 sin190

c)

the equations for the x- component of the position as function of time is

         x = voxt

the equations for the y- component of the position as function of time is

            y = voyt + (1/2)gt2

d)

let horizontal range OB = x

      x - x0 = voxt

      x - (0) = (v0 cos190)(t)   

     x = (8.1 m/s)( cos190)(5 s)  

        = 38.3 m    

e)

let OA = h

y = voyt + (1/2)gt2

y - y0 = (v0 sin210)t + (1/2)gt2

y - (0) = (v0 sin210)t + (1/2)gt2

y  = (v0 sin210)t + (1/2)gt2

let vertical distance y = h

h = (8.1)(sin190) + (1/2)(9.8)(5)2

    = 125.14 m

f)

if h = 10 m

10 = (8.1) (sin190)t + (1/2)(9.8)t2    

4.9t2 + 2.64 t - 10 = 0

we solve the above quadratic equation, we get

      t = 1.184 s

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