The figure shows the launch conditions and an appropriate coordinate system The

# The figure shows the launch conditions and an

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motion. The figure shows the launch conditions and an appropriate coordinate system. The speeds v, v x , and v y are related by the Pythagorean Theorem. 0 v r θ 0 h y x v r j v y ˆ i v x ˆ The squares of the vertical and horizontal components of the object’s velocity are: gh v v y 2 sin 0 2 2 0 2 = θ and 0 2 2 0 2 cos θ v v x = The relationship between these variables is: 2 2 2 y x v v v + = Substitute for v x and v y and simplify to obtain: ( ) gh v gh v gh v v v 2 2 sin cos 2 sin cos 2 0 0 2 0 2 2 0 0 2 2 0 0 2 2 0 2 = + = + = θ θ θ θ 79 •• At 2 1 of its maximum height, the speed of a projectile is 4 3 of its initial speed. What was its launch angle? (Ignore any effects due to air resistance.)
Chapter 3 228 Picture the Problem In the absence of air resistance, the projectile experiences constant acceleration during its flight and we can use constant-acceleration equations to relate the speeds at half the maximum height and at the maximum height to the launch angle θ 0 of the projectile. θ 0 v r 0 y x h 2 1 h The angle the initial velocity makes with the horizontal is related to the initial velocity components. x y v v 0 0 0 tan = θ (1) Substitute y = h and v y = 0 in the equation y a v v y y Δ + = 2 2 0 2 to obtain: 2 0 2 0 gh v h y y = = Δ (2) Write the equation y a v v y y Δ + = 2 2 0 2 for h y 2 1 Δ = : 2 2 2 2 0 2 h g v v h y y y = = Δ (3) We are given that 0 4 3 v v = when h y 2 1 Δ = . Square both sides and express this using the components of the velocity. The x component of the velocity remains constant. ( ) 4 3 2 0 2 0 2 2 2 0 y x y x v v v v + = + (4) where we have used v x = v 0 x . (Equations 2, 3, and 4 constitute three equations in the four unknowns v 0x , v 0y , v y , and h . To solve for any of these unknowns, we first need a fourth equation. However, to solve for the ratio ( v 0y / v 0x ) of two of the unknowns, the three equations are sufficient. That is because dividing both sides of each equation by v 0x 2 gives three equations and three unknowns v y / v 0x , v 0y / v 0x , and h / . 2 x 0 v Solve equation (3) for gh and substitute in equation (2): ( ) 2 2 0 2 0 2 h y y v v v = 2 2 0 2 y y v v = Substitute for v y 2 in equation (4): ( ) 2 0 2 0 2 2 0 2 0 4 3 2 1 y x y x v v v v + = + Divide both sides by v 0 x 2 and solve for v 0y / v 0x to obtain: 1 16 9 2 1 1 2 0 2 0 2 0 2 0 + = + x y x y v v v v 7 0 0 = x y v v
Motion in One and Two Dimensions 229 Substitute for x y v v 0 0 in equation (1) and evaluate θ 0 : ( ) ° = = = 3 . 69 7 tan tan 1 0 0 1 0 x y v v θ 80 •• A cargo plane is flying horizontally at an altitude of 12 km with a speed of 900 km/h when a large crate falls out of the rear-loading ramp. (Ignore any effects due to air resistance.) ( a ) How long does it take the crate to hit the ground? ( b ) How far horizontally is the crate from the point where it fell off when it hits the ground? ( c ) How far is the crate from the aircraft when the crate hits the ground, assuming that the plane continues to fly with the same velocity?

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