6. 1 /3 points | Previous Answers SerCP9 3.P.034. My Notes | Question Part Point

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6.1/3 points |  Previous AnswersSerCP9 3.P.034.My Notes  | Question Part Points Submissions Used A playground is on the flat roof of a city school, 6.2 m above the street below (see figure). The vertical wall of the building is h = 7.50 m high, to form 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 ? = 53.0
6.1/3 points |  Previous AnswersSerCP9 3.P.034.My Notes  |
6.1/3 points |  Previous AnswersSerCP9 3.P.034.My Notes  | Question Part Points Submissions Used A playground is on the flat roof of a city school, 6.2 m above the street below (see figure). The vertical wall of the building is h = 7.50 m high, to form 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 ? = 53.0 Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used A playground is on the flat roof of a city school, 6.2 m above the street below (see figure). The vertical wall of the building is h = 7.50 m high, to form 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 ? = 53.0 A playground is on the flat roof of a city school, 6.2 m above the street below (see figure). The vertical wall of the building is h = 7.50 m high, to form 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 ? = 53.0 Question Part Points Submissions Used A playground is on the flat roof of a city school, 6.2 m above the street below (see figure). The vertical wall of the building is h = 7.50 m high, to form 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 ? = 53.0 degree 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 speed at which the ball was launched. Find the vertical distance by which the ball clears the wall. Did you accidentally divide or take the inverse in your calculation? M Find the horizontal distance from the wall to the point on the roof where the ball lands. Please show work for B and c

Explanation / Answer

since initial V =18.1 m/s

for vertical calculation , let ball is at height h ,time =2.2 s ,initial vertical velocity,u = V*sin(53) =18.1*0.8=14.5 m/s , acceleration due to gravity,a = -9.81 m/s^2

time taken by ball to reach at top point=1.5 s,

topmost height =14.5*1.5-0.5*9.81*1.5^2 =10.7 m

at top point vertical velocity=0, so S in time (t=2.2-1.5=0.7) = 0*t+0.5*g*t^2=0.5*9.81*0.7^2=2.4

so ht of ball =10.7-2.4 =8.3 , so the ball clears the wall by =8.3-7.5=0.8 m

(c) here vertical height of ball =6.2 m, so total cleared by ball =10.7-6.2 =4.5 m

so time = sqrt(4.5/(0.5*9.81)) =0.96 s , total time=1.5+0.96 =2.46 s

total horizontal distance =2.46*18.1*cos(53) =26.72 m ,

distance of point from wall =26.72-24=2.72 m

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