From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 7.50 m/s and angle of 18.0° 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: y0. Assume SI units. Do not substitute numerical values; use variables only.)xi= yi= (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity.vi, x= m/svi, 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: y0 and t. Assume SI units.)x= my= 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 reach a point 10.0 m below the level of launching? s

Initial vertical height of the ball above the ground yi =y0 Final vertical height of the ball is…

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Question

From the window of a building, a ball is tossed from a height y₀ above the ground with an initial velocity of 7.50 m/s and angle of 18.0° 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: y₀. 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	=	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: y₀ 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 reach a point 10.0 m below the level of launching?
s

Expert Solution

Step 1

Initial vertical height of the ball above the ground y_i =y₀

Final vertical height of the ball is y_f=0

Time taken by ball to strike the ground is 5 s

Initial velocity of the ball v₀=7.50 m/s at an angle of 18.0° below the horizontal

$N o w, t h e x a n d y c o m p o n e n t a r e g i v e n b y v_{0 x} = v_{0} \cos θ = 7.50 m / s \times \cos 18 ° = 7.132 m / s v_{0 y} = - v_{0} \sin θ = - 2.31 m / s a . f r o m e q u a t i o n o f m o t i o n w e h a v e, y - y_{0} = v_{0 y} t + \frac{1}{2} a_{y} t^{2} 0 - y_{0} = - 2.31 \times 5 + \frac{1}{2} (- 9.8 m / s^{2}) {(5)}^{2} - y_{0} = - 134.05 y_{0} = 134.05 m b . x - a n d y - c o m p o n e n t o f i n i t i a l v e l o c i t y o f t h e b a l l v_{0 x} = v_{0} \cos θ = 7.50 m / s \times \cos 18 ° = 7.132 m / s v_{0 y} = - v_{0} \sin θ = - 2.31 m / s c . x c o m p o n e n t o f t h e p o s i t i o n o f t h e b a l l a t a n y i n s \tan t o f t i m e i s x = v_{0 x} t = 7.132 m / s \times t y c o m p o n e n t o f t h e p o s i t i o n o f t h e b a l l i s y - y_{0} = v_{0 y} t + \frac{1}{2} a_{y} t^{2} y = y_{0} + v_{0 y} t + \frac{1}{2} (- 9.8 m / s^{2}) {(t)}^{2} = 134.05 m - 2.31 m / s \times t + \frac{1}{2} (- 9.8 m / s^{2}) {(5)}^{2}$

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