A rock is shot vertically upward from the edge of the top of a tall building.The rock reaches its maximum height above the top of the building 1.60 s after being shot ?

Question

Julia Aguilar · Accepted Answer

The initial upward velocity of the rock is #"15.7 m/s"#.

FULL QUESTION A rock is shot vertically upward from the edge of the top of a tall building.The rock reaches its maximum height above the top of the building 1.60 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 8.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building? #stackrel("------------------------------------------------------------------------------------------------------------------------")# At maximum height, the velocity of the rock is equal to zero. Since you know how much time is needed for the rock to reach maximum height, you can say that #underbrace(v_"top")_(color(blue)(=0)) = v_0 - g * t_"up"# This means that you have #v_0 = g * t_"up" = 9.8"m"/"s"^color(red)(cancel(color(black)(2))) * 1.60color(red)(cancel(color(black)("s"))) = color(green)("15.7 m/s"# The distance covered by the rock while it moves upward is #h = v_0 * t_"up" - 1/2 * g * t_"up"^2# #h = 15.7"m"/color(red)(cancel(color(black)("s"))) * 1.6color(red)(cancel(color(black)("s"))) - 1/2 * 9.8"m"/color(red)(cancel(color(black)("s"^2))) * 1.6^2 color(red)(cancel(color(black)("s"^2))) = color(green)("12.6 m")# To find the height of the building, #H#, use the time it takes the rock to reach the ground after reaching its maximum height. In other words, you need to find the distance the rock covers in free fall from its maximum height. SInce it takes the rock a total of 8.00 s to fall to the ground from the moment of its launch, you can say that #t_"down" = t_"total" - t_"up"# #t_"down" = 8.00 - 1.60 = "6.4 s"# This means that you can write #h_"max" = underbrace(v_"top")_(color(blue)(=0)) * t_"down" + 1/2 * g * t_"down"^2# #h_"max" = 1/2 * 9.8"m"/color(red)(cancel(color(black)("s"^2))) * 6.4^2color(red)(cancel(color(black)("s"^2))) = color(green)("200.7 m")# Since maximum height is equal to the height of the building ,#H#, plus the height the rock travels upward, #h#, you can say that #h_"max" = H + h implies H = h_"max" - h# #H = 200.7 - 12.6 = color(green)("188 m")# SIDE NOTE The values are rounded to three sig figs.

Lily Adams · Answer

undefined The time of flight for the rock is 3.20 seconds.