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 5.10 m above the parking lot, and the school building's vertical wall is h-6.50 m high, forming a 1.40 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 each a point vertically above the wall. Due to the nature of his 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 m/s 

(b) Findthe 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.

Answer 1

Given that the ball takes 2.20s to reach a point vertically above the wall which is 24m from point of launching.

Consider the horizontal motion of the ball

There is no acceleration in the horizontal direction, Horizontal velocity remains constant.

d 24m 2.20 + 2120

U = 10.91m/s

u = ucos53 = 10.91m/s

21 = 10.91m)s cos53

(a) ANSWER :
u= 18.1285m/s

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(b)

Find vertical position when time is 2.20s

Consider the vertical motion of the ball

Use formula

s= ut + -at

2769 - pfn =

The ball is going up and g is acting downwards, so put the negative sign

y = usino *t --gt-

y = 18.1285m/s* sin53 * 2.20s - 0.5 * 9.81m/s* (2.20)

y = 18.1285 * sin53 * 2.20 -0.5 * 9.81 * 2.202

y = 31.8517 - 23.7402

${\color{Red} y=8.1115m}$

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Given the height of the school building's vertical wall,

h = 6,50m

Vertical distance by which the ball clears the ball, $c=y-h=8.1115m-6.50m=1.6115m$

(b) ANSWER: ${\color{Red} 1.6115m}$

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(c)

First, find the time when the ball hits the playground

It is time when the vertical height is equal to the playground's height which is 5.10m

Consider the vertical motion of the ball

Use formula

s= ut + -at

2769 - pfn =

The ball is going up and g is acting downwards, so put the negative sign

y = usino *t --gt-

$5.10m=18.1285m/s*sin53*t-0.5*9.81m/s^{2}*t^{2}$

$5.10=18.1285*sin53*t-0.5*9.81*t^{2}$

$5.10=14.4781t-4.905t^{2}$

$4.905t^{2}-14.4781t+5.10=0$

Solve the quadratic equation using a calculator

${\color{Red} t=2.543s}$ (take the larger value, it should come after 2.20s)

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Consider the horizontal motion of the ball

There is no acceleration in the horizontal direction, Horizontal velocity remains constant.

$u_{x}=\frac{X}{t}$

$X=u_{x}*t$

$X=10.91m/s*2.543s$

${\color{Red} X=27.744m}$

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Horizontal distance from the wall to landing point on the playground,

I= X-d=27.744m - 24m = 3.744m

(c) ANSWER: ${\color{Red} 3.744m}$

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