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

The figure shows the launch conditions and an

  • Homework Help
  • Mike_0260
  • 122
  • 100% (2) 2 out of 2 people found this document helpful

This preview shows page 61 - 64 out of 122 pages.

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.)
Image of page 61
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
Image of page 62
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?
Image of page 63
Image of page 64

You've reached the end of your free preview.

Want to read all 122 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes