From the window of a building, a ball is tossed from a height y0 above the groun

Question

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.90 m/s and angle of 21.0° 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: y0.)

(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: y0 and t. Let the variable t be measured in seconds.)

(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?

(E)Find the height from which the ball was thrown.

Explanation / Answer

A)Initial velocity in downward direction = Vsin 21 = 8.9 *sin (21) =3.18 ms-1

Final speed will be = 3.18+ 9.8*6 =62 ms-1

Using newton's equation

V^2 - U^2 = 2g H

H =195 m

Coordinates would be (0,195)

B) Y component would be -3.18 ms-1 and X component is 8.3 ms-1

C) X = 8.3 t

Y = (3.18+9.8t)^2 - 10.11 / 19.6

D) Xmax = 8.3 * 6 = 50 m

E) Y max = 195 m

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