An arrow is shot at an angle of @= 45° above the horizontal. The arrow hits…
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Question “An arrow is shot at an angle of @= 45° above the horizontal. The arrow hits…”
An arrow is shot at an angle of @= 45° above the horizontal. The arrow hits a tree a horizontal distance D = 220 m away, at the same height above the ground as it was shot. Use g = 9.8 m/s for the magnitude of the acceleration due to gravity. Part A Find ta, the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. View Available Hint(s) 190 A ROO? ta Submit
Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. View Available Hint(s) O AQ O ? Submit P Pearson
Answer
Given, –+ angle of projection; O= 45deg Above horizontal and horizontal range: 10 = 220m acceleration because of gravity; g = 90.8 m/s Now, Porn formula of trajectory of is given by y 😡 tane92 2U couro. where yx u is the distance in the y-coordinate distance at the coordinate initial speed projection at x =D=220m. equation , we get om hence hom 0 = 220 tan 45deg – g 2 u2coro. 0=220.1 – 912207 2.4 12 > 220g = 1 = “=1/2rog – u = 46.43 m/s.
DATE 1 PAGE Number. (a) now for x-coordinate, Vai = u oso z 46.43.63 450 Vai = 32.84 m/s. ax = 0. X = 220m. x= Vi l, where 32.84 x 32.84 x 32.84 x 32.84 x 32.84 x 32.84 T= 220 32.84. T = 6.7 seconds spent by the arrow in flight. Hence, tim 6.7 second. (6) Since the tree is dropped from the tree, we need that the apple also falls to the ground. Let’s say that an apple must be dropped from height y. Then, the total time it takes for the apple to reach ground is. Y 7 = Y; +ri- 2gt2 0. = -yto I t = 24, 9.8 9.88 Tz
Ground – then NOW Apple since it takes 6.75m to hit the must also be dropped after the -t =6.7 t=16 7/24 sarrow is shot. Because the height of an apple is not indicated, let’s say y=10m and apple is dropped.
Conclusion
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