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Physics "G" Two Dimensional Motion - by Jon Wisniewski, 2003

1.  Marlene jumps off the edge of a cliff and hits the water 1.5 seconds later, about 4.5 meters from the edge of the cliff.
What height was the cliff? With what speed did she leave the edge?

Here is what you know:

 Horizontal Vertical Dx 4.5 m ? vi n/a 0 m/s vf n/a ? vavg ? n/a a 0 -9.8 m/s/s Dt 1.5 sec 1.5 sec

So, we use the formula Dx = vi(Dt) + 1/2a(Dt)2 to find the height of the cliff. Plug in the numbers so Dx = (0)(1.5 s) + (1/2)(9.8 m/s/s)(1.5 s) = 11.025 m o2 w/ sig. figs 11 m. To find the speed that she left the cliff we use the formula vavg = Dx/Dt.  Plug in the numbers. vavg = (4.5 m)/(1.5 s) = 3 m/s.

2.  Kyle wants to jump into the water from a cliff that is 21 m tall. There are rocks that stick out 3.4 meters from the base of the cliff. What time will he be in the air?  What must his speed be in order to clear the rocks?

Here is what we know:

 Horizontal Vertical Dx 3.4 m 21 m vi n/a 0 m/s vf n/a ? vavg ? n/a a 0 -9.8 m/s/s Dt ? ?

First we must find the time that Kyle is in the air. We use the formula Dx = vi(Dt) + 1/2a(Dt)2 for the vertical components.  Plug in the numbers. 21 m = (0)(Dt) + (1/2)(9.8 m/s/s)(Dt)2 So Dt=2.07 seconds. Now we can find the Speed that he needs to clear the rocks, since time is on both sides. Use the formula vavg = Dx/Dt. vavg = 3.4/2.07 = 1.64 m/s

3. A football leaves the ground with a speed of 27.4 m/s at an angle of 57° above the horizontal.  a) Draw a picture of the initial velocity vector.  b) What is the horizontal velocity?  c) What is the initial vertical velocity component?  d) What time will the ball be in the air?  e) What distance will it go in that time?  f) What is the maximum distance you could make a football go with that speed?

Here is what you know:  Angle of elevation = 57°, speed (v) = 27.4 m/s

 Horizontal Vertical Dx ? ? vi n/a ? vf n/a ? vavg ? n/a a 0 -9.8 m/s/s Dt ? ?

So, first we need to draw the vector and then find the horizontal and vertical components of the velocity.  Using Trigonometry, we can find the velocity components.  The vertical component of velocity = (speed)*Sin(angle of elevation) = (27.4 m/s)Sin(57°) = 22.979 m/s = 23 m/s.  The horizontal component of velocity = (speed)*Sin(angle of elevation) = (27.4 m/s)Sin(57°) = 14.923 m/s = 14.9 m/s.

Now we can find the time that the football spends in the air.  If we set the vf = 0, then the time in the air will be twice that number.  So use the formula vf = vi + a(Dt) when vi = the vertical component of velocity.  Plug in the numbers.  0 = (23 m/s) + (-9.8 m/s/s)(Dt) = 2.3469 s and multiply by 2 to get 4.7 sec.

Since time is on both sides, use the formula vavg = Dx/Dt since a=0 for the horizontal component.  Plug in the numbers. (14.9 m/s) = Dx/ (4.7 sec) = 70.03 m = 70 m.

Now use the Range equation to find the maximum range that the football can travel and use 45° for q.  Range = v2 / g (sin2q) = ( (27.4 m/s)2 / (9.8 m/s/s) ) * (2Sin(45°)) = 76.6 m

4.  A projectile leaves the ground with a speed of 34 m/s at an angle of 37° above the horizontal.  a) What is the initial velocity in vector component notation?  b) What time is the projectile in the air? c) What is its range? d) What is its speed at the highest point? e) What is the velocity of the projectile in vector component notation when it is on the way up at elevation 10 m? f) Speed at elevation 10 m?

Here is what you know: Angle of elevation = 37°, speed (v) = 34 m/s

 Horizontal Vertical Dx ? ? vi n/a ? vf n/a ? vavg ? n/a a 0 -9.8 m/s/s Dt ? ?

First we need to find the vector component form for the velocity.  The horizontal component of the velocity = (speed)*(Cos(angle of elevation)) = (34 m/s)(Cos(37°)) = 27.1 m/s.  The vertical component of the velocity = (speed)*Sin(angle of elevation) = (34 m/s)(Sin(37°)) = 20.5 m/s.  SO the vector is written as follows: 27.1 m/s x + 20.5 m/s y.

Now we can find the time that the projectile is in the air.  Use the formula vf = vi + a(Dt) and set vf = 0 and vi = 20.5 m/s and then multiply your answer by two to get the time.  Plug in the numbers.  0 m/s = 20.5 m/s + (-9.8 m/s/s)(Dt) = 2.09s *2 = 4.1 sec.

We can find the range using the range formula.  Range = v2 / g (sin2q) = ( (34 m/s)2 / (9.8 m/s/s) ) * (2Sin(37°)) = 113 m.

The speed of the projectile at its highest point can be found.  When it is at it's highest point, the vertical component of velocity is = 0, so the speed is merely the horizontal component of velocity which is 27.1 m/s.

The velocity of the projectile at 10 m of elevation is the horizontal component of the velocity (since it doesn't accelerate horizontally) x + the vertical component in the y direction.  To find the vertical component of velocity, we use the formula vf2 = vi2 + 2a(Dx).  vi = 20.5 m/s, a = -9.8 m/s/s and Dx = 10 m.  Plug in the numbers vf = ((20.5)2 + (2)(-9.8 m/s/s)(10 m))1/2 = 14.9 m/s.  Thus in vector component notation the velocity = 27.1 m/s x + 14.9 m/s y.

Now to find the speed when the projectile is 10 m above the ground, just take the square root of the square of the sum of the velocity components.  ie. (x2 + y2)1/2.  Plug in the numbers.  ((27.1 m/s)2 + (14.9 m/s)2)1/2 = 31 m/s.

5.  A motorboat can go 2.4 m/s on a river where the current is 1.8 m/s.  The motorboat must go 240 m upstream, and then back.  What is the speed of the boat with respect to the shore as it travels upstream? Downstream?  How much time does it take it to go upstream? Downstream?

Here's what you know: vmax = 2.4 m/s, current = 1.8 m/s, Dx = 240 m

To find the speed in which the boat travels going upstream and downstream, you merely subtract or add the currents (this is because the vectors are parallel to one another (in the same direction)).  So, the speed in which the boat travels upstream is the difference of the two vectors, which is 2.4 m/s - 1.8 m/s = .6 m/s.  The speed in which it travels downstream is the sum of the two vectors, which is 2.4 m/s + 1.8 m/s = 4.2 m/s.

To find the time that it takes to travel up and downstream, we use the formula vavg = Dx/Dt or Dt = Dx/vavg, where vavg = the speed of the boat traveling up and downstream.  So, the time that it takes to travel upstream is Dt = 240 m / .6 m/s = 400 s.  The time that it takes it to go upstream is Dt = 240 m / 4.2 m/s = 57.1428 s = 57 s

6.  The current in a river 117 m wide is 1.45 m/s, and your boat can go 3.67 m/s.  What time will it take you to cross the river if you point straight across?  What Speed would you go if you pointed straight in to the current?  What angle upstream of straight across must you point your boat to actually go straight across?

Here's what you know:

 Across Down Dx 117 m ? vavg 3.67 m/s 1.45 m/s Dt ? ?

To find the time that it takes for your boat to go straight across the river, use the formula vavg = Dx/Dt or Dt = Dx/vavg.  Plug in the numbers and you get Dt = 117 m / 3.67 m/s = 31.88 s = 31.9 s.

To find the speed in which you would travel if you went directly upstream, you just subtract the velocities (this is because the vectors are parallel to one another (in the same direction)).  So, the speed is 3.67 m/s - 1.45 m/s = 2.22 m/s.

To find the speed that you would travel across the river, you merely use Trigonometry.  Use the formula Tan q = opp. / adj. = Tan q = 1.45 m/s / 3.67 m/s.  So, q = 23.3°.

7.  A ferry boat points upstream at some angle to go straight across a river.  The river current is 1.8 m/s, and it takes the boat 30 seconds to cross the 60 m wide river.  What is the speed of the boat with respect to the shore?  What is the speed of the boat on still water?  What angle upstream of straight across does the boat point?

Here's what you know:

 Across Down Dx 60 m ? vavg ? 1.8 m/s Dt 30 sec 30 sec

To find the speed that the boat travels with respect to the shore use the formula vavg = Dx/Dt = 60 m / 30 s = 2.0 m/s.

To find the speed on the boat in still water use the formula (x2 + y2)1/2.  Plug in the numbers...again... ((2.0 m/s)2 + (1.8 m/s)2)1/2 = 2.69 m/s = 2.7 m/s.

To find the angle that the boat points upstream, you use the formula Tan q = opp. / adj.  Plug in the numbers and Tan q = 1.8 m/s / 2.0 m/s.  So, q = 41.98° = 42°.