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ity ity Horizontal projections Sample Problem 1. ns A ball is thrown horizontally from a building 60.0m high with a speed of 25.0m/s. Find the (a)vertical and horizontal components of the ball's initial velocity, (b) time of flight, (c)distance from the foot of the building where the ball will strike the ground, and (d)velocity when the ball will strike the ground) Given: dy = 60.0m vi = 25.0m/s Solution: (a) x- and y- components of vi: Since the ball is horizontally thrown: Vix = 25.0m/s Viv = 0 (b) First, solve for vfy using vfy2 = Viy + 2ady, since Viy = 0: Vfy = 2ady = V2(-9.8m/s2)(-60.0m) = -34.29m/s To get t: Vfy = Viy + at t = Ofy - Viy _-34.29m/s = 3.50s a -9.8m/s (c) Range: dx = Vixt = (25m/s)(3.50s) = 87.50m (d) Get the resultant of the components vfx and vry:1. A stone is dropped from the top of a building and falls freely from rest. What is its position (height) and its velocity after 55 and 65? 2. A boy threw a ball upward with an initial velocity of 40 m/s and was able to catch it before it reached the ground on its return. a) What was the ball's velocity after 'ls? b) What was the ball's velocity after 25? c) What was its displacement after \"Is? all How long did it take the ball to reach its maximum height? e) How far was the maximum height from the starting point? 3. A ball is thrown horizontally from a building 60.0m high with a speed of 35.0m/s. Find the (a)vertical and horizontal components of the ball's initial velocity, (b) time of flight, (c)distance from the foot of the building where the ball will strike the ground, and (d) velocity when the ball will strike the ground Question: When can be use negative or positive acceleration/gravity? Explain Solutions: d. t from origin to maximum height a. vafter Is D Note: An object thrown upward will user Dy = Vi- gt reach a certain point (maximum by= 20 m/s-(9.8 m/s )(1.0s) height) where it will momentarily by = 20 m/s - 9.8 m/s stop and then start to move By = 10.2 m/s downward. At this point, vy = 0. b. v after 2s Use: vi = Vi - gt use. By = m - gt Manipulating the equation, by= 20 m/s -(9.8 m/s )(2.0s) vy = 20 m/s - 19.6 m/s t = Vi - Uf 9 vy= 0.4 m/s t=20 m/s - 0 c. d after Is 9.8 m/s- use: t = 2.04 s d = vit -: d= 20m/s(1.0s) - (9.8 m/s )(1.0s)2 e. d from origin to maximum height 2 use: d = 20m - 4.9 m d = vit - 9 d = 15.1 m d = 20m/s (2.04 s) - (9.8 m/s2)(2.04s) (positive sign denotes that the object is 2 moving upward) d = 40.8 m - 20.40 m d = 20.4 mVf = Vvfx 2 + Vfy 2 = (25m/s)2 + (-34.29m/s)2 = 42.44m/s -34.29 For the direction: 0 = tan-1 = tan -1 = - 53.910 25 vf = 42.44m/s, 53. 91 South of EastSample Problem 1. A stone is dropped from the top of a building and falls freely from rest. What is its position (height) and its velocity after 1.0s, 2.0s, and 3.0s? To find for the position(height), we will c. Find h after 3.0s h = 0(3.0s) - (9.8 m/s?)(3.0s)2 use: h = vit-9 h = 0 - 88.2 m Know that initial velocity (vi) = 0 2 a. Find h after 1.0s h = - 44.1 m or 44.1 m downward h = 0(1.0s) - (9.8 m/s )(1.0s) To find for velocity (vr), we will use: v/ = VI - gt h = 0-9.8 m a. Find vyafter 1.0s h = - 4.9 m vy = 0- (9.8 m/s )(1.0s) (negative sign denotes downward direction) vy = - 9.8 m/s b. Find h after 2.0s b. Find vy after 2.0s h = 0(2.0s) - (9.8 m/s )(2.0s)2 vy = 0 - (9.8 m/s ) (2.0s) My= - 19.6 m/s h =0-39.2m c. Find vy after 3.0s vy = 0 - (9.8 m/s2)(.0s) h = - 19.6 m or 19.6 m downward vy = - 29.4 m/s Sample Problem 2. A boy threw a ball upward with an initial velocity of 20 m/s and was able to catch it before it reached the ground on its return. a. What was the ball's velocity after Is? b. What was the ball's velocity after 2s? c. What was its displacement after Is? d. How long did it take the ball to reach its maximum height? e. How far was the maximum height from the starting point? Given: vi = 20 m/s Find: a. v after Is b. v after 5s c. d after Is d. t from origin to maximum height e. d from origin to maximum height