From the window of a building, a ball is tossed from a height y_0 above the grou

Question

From the window of a building, a ball is tossed from a height y_0 above the ground with an initial velocity of 8.10 m/s and angle of 22.0 degree below the horizontal. It strikes the ground 6.00 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.) x_i = 0 y_i = h (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. v_i, x = m/s v_i, y = m/s (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.) x = m y = m (d) How far horizontally from the base of the building does the ball strike the ground? m (e) Find the height from which the ball was thrown. m (f) How long does it take the ball to reah a point 10.0 m below the level of launching? s

Explanation / Answer

Given,

Initial velocity, vo = 8.10 m/s

Angle, = 22°

Time, t = 6 sec

a) Initial coordinates of the ball,

x = 0 m,

y = voyt + (1/2)gt2 = (3.03 x 6) + (0.5 x 9.8 x 36) = 194.58 m

coordinate : (x,y) = (0, 194.58) m

b) x-component of the initial velocity of the ball,

        v0x = v0 cos22° = 7.51 m/s

y-component of the initial velocity of the ball,

        v0y = v0 sin22° = 3.03 m/s

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

         x = voxt = 7.51 x 6 = 45.06 m

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

            y = voyt + (1/2)gt2 = (3.03 x 6) + (0.5 x 9.8 x 36) = 194.58 m

d) let horizontal range OB = x

      x - x0 = voxt

      x - (0) = (v0 cos22°)(t) = (8.10)( cos22°)(6) = 45.06 m

e) let OA = h

y = voyt + (1/2)gt2

y - y0 = (v0 sin22°)t + (1/2)gt2

y - (0) = (v0 sin22°)t + (1/2)gt2

y  = (v0 sin22°)t + (1/2)gt2

let vertical distance y = h

h = (8.1)(sin22°) + (1/2)(9.8)(6)2 = 179.43 m

f) if h = 10 m

10 = 8.10 x (sin22°)t + (1/2)(9.8)t2    

4.9t2 + 3.03t - 10 = 0

we solve the above quadratic equation, we get

      t = 1.15 s

Please rate my answer if you find it helpful, good luck...

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.