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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 6.10 m above the parking lot, and the school building's vertical wall is h = 7.40 m high, forming a 1.30 m high railing around the playground. The ball is launched at an angle of 8 = 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-including answers submitted in WebAssign-in your calculations.) (a) Find the speed (in m/s) at which the ball was launched. 18.1 m/s (b) Find the vertical distance (in m) by which the ball clears the wall. 0.73 ✓ m (c) Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands. 2.68 m (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(8) = 1 + tan²(€).) 47.7 ⚫ above the horizontal (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? × The approach you use should be identical to part (c), only now the initial angle is the value found in part (d). m

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 6.10 m above the parking lot, and the school building's vertical wall is h = 7.40 m high, forming a 1.30 m high railing around the playground. The ball is launched at an angle of 8 = 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-including answers submitted in WebAssign-in your calculations.) (a) Find the speed (in m/s) at which the ball was launched. 18.1 m/s (b) Find the vertical distance (in m) by which the ball clears the wall. 0.73 ✓ m (c) Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands. 2.68 m (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(8) = 1 + tan²(€).) 47.7 ⚫ above the horizontal (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? × The approach you use should be identical to part (c), only now the initial angle is the value found in part (d). m

Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter3: Vectors
Section: Chapter Questions
Problem 42PQ: The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F1 =...
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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 6.10 m above the parking lot, and the school building's vertical wall is h = 7.40 m high, forming a 1.30 m high railing around the playground. The ball is launched at an angle of 8 = 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-including answers submitted in WebAssign-in your calculations.) (a) Find the speed (in m/s) at which the ball was launched. 18.1 m/s (b) Find the vertical distance (in m) by which the ball clears the wall. 0.73 ✓ m (c) Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands. 2.68 m (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(8) = 1 + tan²(€).) 47.7 ⚫ above the horizontal (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? × The approach you use should be identical to part (c), only now the initial angle is the value found in part (d). m
Transcribed Image Text: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 6.10 m above the parking lot, and the school building's vertical wall is h = 7.40 m high, forming a 1.30 m high railing around the playground. The ball is launched at an angle of 8 = 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-including answers submitted in WebAssign-in your calculations.) (a) Find the speed (in m/s) at which the ball was launched. 18.1 m/s (b) Find the vertical distance (in m) by which the ball clears the wall. 0.73 ✓ m (c) Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands. 2.68 m (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(8) = 1 + tan²(€).) 47.7 ⚫ above the horizontal (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? × The approach you use should be identical to part (c), only now the initial angle is the value found in part (d). m
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