A ball is launched from the top of an 24.8 m tall building at an angle = 26.0° above horizontal with a speed of 2.85 m/s. What is the ball's horizontal distance (in m) from the building at the time it reaches the ground? Use g = 9.81 m/s?). (These are my steps but I don’t know what to do next. The outcome should be 6.1. I did 2.13*2.562 and got 5.46. Don’t know what I did wrong)

Physics for Scientists and Engineers, Technology Update (No access codes included)
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ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter1: Physics And Measurement
Section: Chapter Questions
Problem 1.6OQ
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A ball is launched from the top of an 24.8 m tall building at an angle = 26.0° above horizontal with a speed of 2.85 m/s. What is the ball's horizontal distance (in m) from the building at the time it reaches the ground? Use g = 9.81 m/s?). (These are my steps but I don’t know what to do next. The outcome should be 6.1. I did 2.13*2.562 and got 5.46. Don’t know what I did wrong)
2.85 Cos (26) = 2.562
2.85 Sin (26) −1.2494
2
24.8- (1.2494)T + ½ (98|) 1
4.912 + 1.2494_T-24.8 + >o
4.912 + 1.2494 + - 24. 8 = 0
Use quadratic
lavation
out come from equestion.
+ - - 2, 3 8
12 : 2.13
Transcribed Image Text:2.85 Cos (26) = 2.562 2.85 Sin (26) −1.2494 2 24.8- (1.2494)T + ½ (98|) 1 4.912 + 1.2494_T-24.8 + >o 4.912 + 1.2494 + - 24. 8 = 0 Use quadratic lavation out come from equestion. + - - 2, 3 8 12 : 2.13
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