Children playing in a playground on the flat roof of a city school lose their ba

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 5.00 m above the parking lot, and the school building's vertical wall is

h = 6.50 m

high, forming a 1.50 m high railing around the playground. The ball is launched at an angle of

= 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.

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

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

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() = 1 + tan2().)

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

Explanation / Answer

vo = initial velocity of launch of the ball

consider the motion along the horizontal direction

Vox = initial velocity = vo Cos53

X = horizontal displacement = d = 24 m

t = time taken = 2.20 s

using the equation

X = Vox t

24 = (vo Cos53) (2.20)

vo = 18.13 m/s

consider the motion in vertical direction :

Voy = initial velocity = vo Sin53 = (18.13) Sin53 = 14.5 m/s

a = acceleration = - 9.8 m/s2

Y = vertical displacement = ?

t = time of travel = 2.20 s

using the equation

Y = Voy t + (0.5) a t2

Y = (14.5) (2.20) + (0.5) (- 9.8) (2.20)2

Y = 8.2 m

h = height of the wall = 6.50 m

vertical distance by which the ball clears the wall is given as

vertical distance = Y - h = 8.2 - 6.50 = 1.7 m

b)

let the time of travel be t' for the ball hit some point on the roof

consider the motion in vertical direction :

Voy = initial velocity = vo Sin53 = (18.13) Sin53 = 14.5 m/s

a = acceleration = - 9.8 m/s2

Y = vertical displacement = 6.5

t' = time of travel = ?

using the equation

Y = Voy t' + (0.5) a t'2

6.5 = (14.5) t' + (0.5) (- 9.8) t'2

t' = 2.4 sec

consider the motion along the horizontal direction

Vox = initial velocity = (18.13)Cos53 = 10.91 m/s

X' = horizontal displacement = ?

t = time taken = 2.4 s

using the equation

X' = Vox t

X' = (10.91) (2.40)

X' = 26.2 m

X = d = distance upto wall = 24 m

horizontal distance = X' - X = 26.2 - 24 = 2.2 m

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