Question

Children playing in a playground on the flat roof of a city school lose their ball to the parking lot below. One of the teachers kicks the ball back up to the children as shown in the figure below. The playground is 5.50 m above the parking lot, and the school building's vertical wall is

h = 6.90 m

high, forming a 1.40 m high railing around the playground. The ball is launched at an angle of

θ = 53.0°

above the horizontal at a point

d = 24.0 m

from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

a) Find the speed (in m/s) at which the ball was launched.

(b) Find the vertical distance (in m) by which the ball clears the wall.

(c)Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands.

(d)What If? If the teacher always launches the ball with the speed found in part (a), what is the minimum angle (in degrees above the horizontal) at which he can launch the ball and still clear the playground railing? (Hint: You may need to use the trigonometric identity

sec²(θ) = 1 + tan²(θ).)

(e)What would be the horizontal distance (in m) from the wall to the point on the roof where the ball lands in this case?

Answer 1

(a) ball takes 2.2 s to reach a point vertically above the ball.

horizontal distance of that point is d = 24 m

horizontal component of initial velocity ux = ucos(53)

In horizontal direction there is no acceleration therefore,

d = ucos(53) x t

24 = ucos(53) x 2.2;

initial velocity u = 18.13 m/s

(b) using equation of motion in y direction

y = usin(53) x t - 0.5 x g x t^2

at t= 2.2 s

y = 8.114 m

therefore, the the ball clears the wall by h'

h' = y - h = 8.114 - 6.9

h' = 1.214 m

(C) using equation of motion in y direction

y = usin(53) x t - 0.5 x g x t^2

find time at which y = h - a = 6.9 - 1.4

y = 5.5 m

5.5 = 18.13sin(53) x t - 0.5 x 9.8 x t^2

4.9 t^2 - 14.479 t + 5.5 = 0

solve the quadratic equation

t = 2.507 s

now calculate D at t = 2.507 s

D = ucos(53) x t

D = 18.13 x cos(53) x 2.507

D = 27.354 m

distance from the wall = D - d = 27.354 - 24

= 3.354 m

(D) for just clear the ball;

y = 6.9 m and d = 24 m

6.9 = (18.13sin)t - 0.5gt^2 ..........................(1)

also 24 = (18.13cos)t .............................(2)

substituting the value of t from equation (2) to equation (1)

6.9 = (18.13sin) x (24/18.13cos) - 0.5g x (24/18.13cos)^2

6.9 = 24tan - 8.595sec^2

6.9 = 24tan - 8.595 x (1 + tan^2)

8.595 tan^2 - 24 tan + 15.495 = 0

solve the abobe equation

tan = 1.0134 , 1.778 (taking the lowest value for minimum angle)

tan = 1.0134

minimum angle = 45.38 degrees

(E) calculate time from from equation (t)

6.9 - 1.4 = (18.13sin45.38)t - 0.5gt^2

5.5 = 12.905t - 4.905x t^2

t =2.096 s

now D = (18.13cos45.38)t

D = (18.13cos45.38) x 2.096

D = 26.692 m

distance from the wall = 26.692 - 24 = 2.692 m