Question

1. Motion in two dimensions.
   1. A plane is flying due south at 560 km/h. A wind blows the plane from east to west at 95 km/hr (yikes!) Find the plane’s velocity with respect to the ground.
   2. A projectile is shot from the edge of a cliff 258 m above ground level with an initial speed of 95.0 m/s at an angle of 22 degrees with the vertical. Find the time taken by the projectile to hit the ground.
   3. Now find the distance away from the base of the cliff the projectile will hit
   4. Then find the horizontal and vertical components of its velocity just before it hits the ground
   5. Next find, the overall velocity and the angle of the velocity vector with the horizontal
   6. Finally, calculate the maximum height above the cliff top the projectile reaches.
   7. Could the projectile travel any farther? Defend your answer.
   8. BONUS: A boat is moving through still water at a speed of 1.75 m/s. The engine propels the boat perpendicular to the banks. However, there is a strong current of 5.63 m/s that also impacts the boat, and a person is walking aft on the deck of the boat (towards the rear) at a speed of 0.80 m/s. What is the person’s velocity relative to the bank (include an angle)? If the river is 20 m wide, and the boat is 4 m long, will the person reach the back before the boat makes it to shore? How far downstream will the boat drift before it makes it across?

Answer 1

a) Let S-N direction be denoted by (j) and W-E direction be denoted by (i).

Hence the initial velocity of the plane, u = 560(-j)

Velocity of the blowing wind, w = 96(-i)

Hence the final velocity of the plane wrt ground,v = where arccos term represents the deviation of the plane from (-j) direction.

Therefore v = 568.169 km/h, with a deviation of 9.727^o from -j direction towards west.

In component form v = 96 (-i) + 560 (-j)

b) Let height of the cliff be h = 258m

v be the launch velocity = 95m/s

g = 9.8m/s²

Using equation of motion h = ut + 0.5at²

Substituting the corresponding values with appropriate sign

-258 = 95cos(22^o) t + (0.5)(-9.8)t²                                  u = 95cos(22)= 88.0824m/s is the vertical component of launch velocity.

Solving the quadratic for t we get

t = 20.5395 secs

c) for Range, R of the projectile we consider the horizontal component of the launch velocity ,x= 95sin(22) = 35.587m/s

time of flight t = 20.5395

R = xt = 35.587*20.5395 = 730.952 m

d) As there is no force acting in the horizontal direction the velocity x will remain same upon impact.

Thus the horizontal velocity upon impact = x = 35.587 m/s

For vertical component we use equation of motion =

= final vertical velocity upon impact

= Vertical velocity of projection = 88.0824m/s

a = -9.8 m/s²

t = time of flight = 20.5395 sec

= 88.0824 - 9.8*20.5395 = 113.2046 , in the downward direction.

e) Final velocity can be calculated using conservation of mechanical energy

Initial mech. energy = mgh + 0.5mv²

Final mech. energy = 0.5mz²

m(-9.8*258 + 0.5*95²) = 0.5mz²

z =

z = 62.99m/s