1.) A stone is dropped into a river from a bridge at heighth above the water.Ano
ID: 1759909 • Letter: 1
Question
1.) A stone is dropped into a river from a bridge at heighth above the water.Another stone is thrown vertically down at time t after the first is dropped.Both stones strike the water at the same time. What is the initialspeed of the second stone? Give your answer in terms of thegiven variables and g. I need a formula for the initial speed of the second stone invariables t,h, and g. 2.) A steel ball is dropped is dropped from a building's roofand passes a window, taking 0.12s to fall from the top to thebottom of the window, a distance of 1.20m. It then falls to asidewalk and bounces back past the window, moving from bottom totop in 0.12s. Assume that the upward flight is an exact reverse ofthe fall. The time the ball spends below the bottom of the windowis 2.35s. How tall is the building? Much appreciated. (I rate superbly to those whole answerclearly) 1.) A stone is dropped into a river from a bridge at heighth above the water.Another stone is thrown vertically down at time t after the first is dropped.Both stones strike the water at the same time. What is the initialspeed of the second stone? Give your answer in terms of thegiven variables and g. I need a formula for the initial speed of the second stone invariables t,h, and g. 2.) A steel ball is dropped is dropped from a building's roofand passes a window, taking 0.12s to fall from the top to thebottom of the window, a distance of 1.20m. It then falls to asidewalk and bounces back past the window, moving from bottom totop in 0.12s. Assume that the upward flight is an exact reverse ofthe fall. The time the ball spends below the bottom of the windowis 2.35s. How tall is the building? Much appreciated. (I rate superbly to those whole answerclearly) Much appreciated. (I rate superbly to those whole answerclearly)Explanation / Answer
I am still working on #1, but here is #2... -- #2. d = vt + 0.5at2 You can use this motion equation to find the velocity of theball as soon as it reaches the top of the window, when it ison the way down. 1.2 = v(0.12) +0.5(9.8)(0.12)2 v = 9.41 m/s (This is also the velocity that the ball has when it reachesthe top of the window on it's way back up.) -- When the ball reaches its highest point (the top of thebuilding), its velocity will be zero, so use this equation to findthe time it takes to reach the top of the building... Vf = Vi + at 0 = 9.41 + (-9.8)(t) t = 0.96 seconds -- Then, use this equation to find the distance from the top ofthe window to the top of the building... d = vt + 0.5at2 d = (9.41)(0.96) +(0.5)(-9.8)(0.962) d = 4.51 m --Then, use this equation to find the velocity of the ball at thebottom of the window. -- Then, use this equation to find the distance from the top ofthe window to the top of the building... d = vt + 0.5at2 d = (9.41)(0.96) +(0.5)(-9.8)(0.962) d = 4.51 m Vf = Vi + at Vf = 9.41 + 9.8(0.12) Vf = 10.586 m/s -- Since the up and down trips are exactly the same, we know thatthe ball spends 1.175 seconds between passing the window andhitting the ground (2.35 seconds / 2). So, use this equation to find the distance below thewindow... d = vt + 0.5at2 d = (10.586)(1.175) +0.5(9.8)(1.1752) d = 19.21 m -- So, now you have the distance from the bottom of the window tothe ground and the distance from the top of the window to the topof the building, and the size of the window, so add them up and youhave the height... (1.2) + (19.21) + (4.51) = 24.92m -- So, now you have the distance from the bottom of the window tothe ground and the distance from the top of the window to the topof the building, and the size of the window, so add them up and youhave the height... (1.2) + (19.21) + (4.51) = 24.92m
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