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A playground is on the flat roof of a city school, 5.5 m above the

0 votes

February 18th, 2013 12:42
in Advanced Physics by anwesha (Air University (AL), Engineering)

A playground is on the flat roof of a city school, 5.5 m above the street below. The vertical wal?

"A playground is on the flat roof of a city school, 5.5 m above the street below. The vertical wall of the building is h = 6.80 m high, forming a 1.3-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it 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.

 

(a) Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.)

 

 

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

 

 

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

6 Answers
0 votes

March 6th, 2013 12:02
eha (Shree Ram Swarup College of Engineering & Management , Electrical Engineering)
"a) Horizontal distance x = 24 = u*2.2 cos 53 u = 18.13 m/s b) The vertical distance travelled in this time is ut – ½ a t^2 y = 18.13 *2.2 sin 53 – 4.9*2.2*2.2 = 8.14 m The distance above the wall is 8.14 – 6.8 = 1.34 m c) The vertical distance of the play ground from the ground is 5.5 m The time to fall on the play ground is found from 5.5 = 18.13 *t* sin 53 – 4.9t^2 t = 0.45 s and this is neglected since it is less than 2.2 s The other value is 2.51s The horizontal distance traveled in this time is 18.13 *2.51cos 53 = 27.39 m"

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May 15th, 2013 10:15
cool_bre (Anna University of Technology, Databases)
"arranged the actual equation of movement inside straight since b=3 in the surface of the play ground to the soccer ball b(to)=-half a dozen+v0*hell(fifty three)*t-.your five*being unfaithful.eighty one*testosterone levelsenqa_two were considering the fact that 26=v0*cos(liii)*only two.30 work out pertaining to v0 20.637 meters/utes talk to Y(big t) to find out in the event the basketball is definitely over this rampart ful(only two.2)=-six+19.637*wickedness(53)*only two.3-.v*nine.eighty one*ii… b(a couple of.ii)=several.762 l, which can be three.762 m over this wall structure. Why don't we uncover t when Y(t)=3 There'll be 2 root base, operate the better root. Small price of testosterone levels is the place the soccer ball actually reaches this aircraft on the play ground on the rise. testosterone levels=2.753 a few moments As meter the actual baseball features sailed times(3.753)=30.your five l or even 27.5-26 from the rampart ix.5 michael "  

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May 16th, 2013 08:43
agarkar (Punjab Engineering College, Mechanical Engineering)
"Your initial Y part is a absolutely nothing ahead of the soccer ball is definitely placed. I can find out this y simply part of your initial speed by taking the original acceleration of (a new) and developing this aside sin involving fifty three certifications. I buy 14.218...l/s. Given that I'm searching for the particular ymca ingredient that matches to help ten length of twenty-four michael, I exploit testosterone levels = a couple of.24: yf = zero + (age 14.218)(only two.per day) + .5 various(- being unfaithful.8)(2.per dayenqa_2) = 8.2628...m "  

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May 27th, 2013 14:06
shaje_69kinky (Amrita Vishwa Vidyapeetham, Computer networks)
"The initial y portion is at zero before the ball is thrown. I can find out the y portion of the initial velocity by taking the initial velocity of (a) and multiplying it by sine of 53 degrees. I get 14.218...m/s. Since I'm trying to find the y component that corresponds to x distance of 24 m, I use t = 2.24:
 
yf = 0 + (14.218)(2.24) + .5(- 9.8)(2.24^2)
 
= 7.2628...m
 
"

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May 31th, 2013 07:26
radhak (Ambedkar University, Delhi, Operating systems)
(hundred) In the event that our train of thought is accurate for the latter parts, and then that is certainly full. Because I am just having problems along with (c). My personal primary imagined is applying that time where to = only two.xxiv for preliminary point, to ensure the times in addition to y simply components turn out to be xi and yi in equations. We are...just mixed up. In the event that using yf = yi + vyit + .5a(tenqa_2), the new yi is the previous(a) yf (that's several.over 250...l). The brand new vyi will be the aged vyf, which is often worked out by simply multiplying -9.8m/(Senqa_two) aside a couple of.24s. Some time it takes from the retaining wall in order to showing up in top is actually challenging nevertheless, that is certainly some tips i are unable to figure out. I guess this vyf = vyi + ayt may do the job, plainly believe the last speed (after the item visitors the cover) may be the bad on the master preliminary speed (way back once the baseball seemed to be thrown). Although, which is the job. The roof can be an impediment--something which impedes the most common parabolical journey. Basically desired the rate upon top-to be able to-musket ball get in touch with, I might need the time, as well as, very well, it is precisely what Furthermore , i will need. ...this can be moving in a new revolve around my personal brain, i possibly need to land one more equality into your combine. XDDD  

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June 1st, 2013 12:04
saryu (AMET University, Mechanical Engineering)
a new) Can it acquire ii.per day mere seconds to achieve the actual wall, or maybe a couple of.something like 20? If you don't wish to record decimal fraction areas, only say in which five_ten = twenty-four/2.xx in addition to keep it this way. Which will help save far more some time to problems when compared with rescue the many decimals.

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