Please see att From the window of a building, a ball is tossed from a height \( y_0 \) above the ground with an initial velocity of 7.80 m/s and angle of 21.0° 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.)\[x_i = \_\_\_\]\[y_i = \_\_\_\]**(b)** With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity.\[v_{i,x} = \_\_\_ \text{ m/s}\]\[v_{i,y} = \_\_\_ \text{ m/s}\]**(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.)\[x = \_\_\_ \text{ m}\]\[y = \_\_\_ \text{ m}\]**(d)** How far horizontally from the base of the building does the ball strike the ground?\[\_\_\_ \text{ m}\]**(e)** Find the height from which the ball was thrown.\[\_\_\_ \text{ m}\]**(f)** How long does it take the ball to reach a point 10.0 m below the level of launching?\[\_\_\_ \text{ s}\]

Answered: Image /qna-images/answer/d73a0a22-0c92-4736-8bdd-edbba2565cde.jpga)When the base of the building is at the origin, the x-coordinate of the ball’s initial position will be zero.Since the ball is at a height of the y0 above the ground the y-coordinate of the initial position of the ball will be y0.b)Calculate the x-component of the initial velocity.Calculate the y-component of the initial velocity.c)Since the velocity along the horizontal direction remains the same, the acceleration along the horizontal direction will be zero.Calculate the x-component of the position.Calculate the y-component of the position.