Question

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

Answer 1

Given, - - + angle of projection; O= 45° Above the horizontal and horizontal range : 10 = 220m acceleration due to gravity ; g = -9.8 m/s Now, Porn equation of trajectory of given by , y : x tane- 92 2U couro - where y x u is the is the is the distance in y-coordinate distance in a coordinate initial speed of projection at x =D = 220m. equation , we get om hence hom 0 = 220 tan 45° - g 2 u²coro. 0=220.1 - 912207 2.4 12 > 220g = 1 = "=1/2rog - u = 46.43 m/s.

/ DATE 1 PAGE No. (a) now for x-coordinate, Vai = u oso z 46.43.63 450 Vai = 32.84 m/s. ax = 0 X = 220 m. So total time of flight is , spent T arrow Ý total time in the air. x= Vi ł where by 220 = 32.84 x T T= 220 32.84 T = 6.7 Seconds spends by the arrow in the air is Hence tim 6.7 seconds. (6) since the tree hits the tree al ground, then we require that the apple hit the ground at the same time. let be apple is to be dropped from height y then total time taken by apple to hit the ground is . y 7 = Y; +ri - 2gt² 0 = -y to I t = 24 9.8 9.88 tz

ground - then NOW apple since takes 6.75 to hit the must be dropped at after the - t = 6.7 t = 16 7 /24 s arrow is shot . Since height of apple is not given lat suppose y=10m then apple is dropped is so, time after 6o7- t = 2X10 - T5.2757 then is So it height at from apple is to be dropped it must be dropped after at 5.27 spaarrow shot.