Mu the super cow sees the Moon ahead in the horizon (moon rise) and tries to jump over the Moon from atop a 14.7 m tall barn with an initial velocity of 78.4 m/s [Up 60o Fwd] . The acceleration due to gravity is 9.8 m/s2 [Down]. Determine her velocity at 4.9 s. Determine when and where she lands. Theta the bird swoops counter-clockwise, accelerating from 29.4 m/s [East 45o North] with an acceleration of "3g's" [West 45o North] for 3.14 s. Determine her displacement.

Nu the horse accelerates from an initial velocity of 20.0 m/s [W 30o N] for 2.5 s ending up with a displacement of 40 m [W 15o S]. Determine its final velocity

Epsilon the hen has an acceleration of 3 g's [N 15o E] for 2.5 s from an initial velocity of 18 km/h [E 30o N]. Determine the final velocity

Solution By Steps for Mu Step 1: Calculate Initial Velocity Components . Given: Vo = 78.4m/s, 0 = 60 . Horizontal Velocity (Vx0): Vx0 = Vo cos(0) = 78.4 cos(60) = 39.2m/s . Vertical Velocity (Vy0): Vyo = Vo sin(0) = 78.4 sin(60 ) = 67.9m/s Step 2: Determine Velocity at 4.9 seconds . Given: t = 4.9 s, g = 9.8 m/2 . Vertical Velocity (Vy): Vy = Vyo - g x t = 67.9 - 9.8 x 4.9 ~ 19.9m/s . The horizontal velocity (Vx) remains constant at 39.2m/s.Step 3: Determine Time of Flight . Using: Y=Yy X t_% X gXx t? +y0, where p = 14.7m . . 1 2 . - Quadratic Equation: = 0 leads to 2 X 9-8 X 1" +67.9 x +14.7=0 . Solving for gives gt =~ 14.1seconds Step 4: Determine Landing Distance . Distance: Distance = v, x Time of Flight = 39.2 x 14.1 ~ 551.5m Final Answer Mu, the super cow, will have a vertical velocity of approximately 19.9m/s and a constant horizontal velocity of 39-2m/s at 4.9 seconds. She lands after approximately 14.1 seconds and covers a distance of approximately 551.5 meters. Key Concept Projectile Motion Key Concept Explanation Projectile motion is a form of motion experienced by an object or particle that is thrown near the Earth's surface and moves along a curved path under the action of gravity only. This motion can be divided into two components: horizontal and vertical. The horizontal motion occurs at a constant velocity, while the vertical motion is influenced by gravity, causing acceleration downwards. By breaking down the motion into these two components, we can analyze and predict the trajectory of the projectile. Solution By Steps for Theta Step 1: Calculate Initial Velocity Components . Given: Vn = 29.4m/s at 45. East of North . Components: . Vx0 = 29.4 sin (45) ~ 25.5 m/s East . Vyo = 29.4 cos(45) ~ 14.7m/s North Step 2: Calculate Acceleration Components . Given: Acceleration of "3g's" = 3 x 9.8m/s at 45. West of North . Components: . ax = -3g sin(45) ~ -25.5 m/s' West . ay = 3g cos(45) ~ 14.7 m/s NorthStep 4: Calculate Displacement Components . Displacement: . de = Vx X t+ , x a x th = 25.5 x 3.14 + 5 x (-25.5) x (3.14)