Pre-Quiz 3.2 .:. Go Up

Projectile Motion

1-3: A ball rolls off the edge of a cliff.  The instant it leaves the edge, it has an initial horizontal velocity of 17.3 m/s, and it strikes the ground after 1.56 seconds.

 1. How high is the cliff? (11.9 m)     2. How far from the base of the cliff does the ball land? (27.0 m)       3. What is the speed of impact? (23.1 m/s) Horizontal s u = 17.3 m/s v a = 0 (horizontally) t = 1.56 s Vertical s u = 0 (purely horizontal) v a = -9.81 m/s/s (on earth) t = 1.56 s

Solution:

So in the vertical direction you can use v = u + at (Not in the packet - you must memorize it!!!) to find the final vertical velocity:

v = 0 + (-9.81)(1.56)  = -15.3036 m/s

And you can find the vertical displacement using s = ut + 1/2at2:

s = 0 + 1/2(9.81)(1.56)2 = 11.936 m (This is the answer to 1)

In the horizontal direction you can find the displacement using s = ut + 1/2at2: - (a = 0), so this becomes s = ut

s = ut = (17.3)(1.56) = 26.988 m (Answer to 2)

As for the speed of impact, the velocity of impact is 17.3 m/s in the x direction, and -15.3036 m/s in the y, the speed is the hypotenuse of this:

= sqrt(17.32 + 15.30362) = 23.097 m/s (The answer to 3)

Questions 1, 2 and 3 are to see if you can solve a cliff problem:

4-5: A ball is launched at 43.2 m/s at an angle of 25.2o above horizontal on a level field.

 4. How far does the ball go before striking the ground? (147 m) Horizontal s = u = 39.089 m/s (see below) v = 39.089 m/s a = 0 (horizontally) t = 3.7500 s Vertical s = 0 u = 18.394 m/s v = -18.394 m/s a = -9.81 m/s/s (on earth) t = 3.7500 s

Solution:

If you get an angle magnitude vector to start with, you have to break it into components.  If the 25.2o is with the x-axis, then it is the trig angle, and the initial x and y components are:

Vx = 43.2cos(25.2o) = 39.089 m/s

Vy = 43.2sin(25.2o)  = 18.394 m/s

We know in the vertical direction two other things, s = 0 (The field is level), and that v = -18.394 m/s (it hits the ground going down as fast as it was going up to begin with)

It is a simple matter to use v = u + at (if you remember it) to find time in the vertical direction:

-18.394 = 18.394 + (-9.81 m/s/s)t

t = 3.7500 s

And finally to find the horizontal displacement:

s = ut + 1/2at2

a = 0, so

s = ut = (39.089 m/s)(3.7500 s)

s = 146.581 = 147 m

 5. There is a very tall wall 130. m away.  With what velocity does the ball strike the wall?  (Express your answer as an angle-magnitude vector – draw a picture) (41.6 m/s @ 20.0o below horizontal) Horizontal s = 130. m u = 39.089 m/s (see below) v = 39.089 m/s a = 0 (horizontally) t = 3.3258 s Vertical s = ? u = 18.394 m/s v = ? a = -9.81 m/s/s (on earth) t = 3.3258 s

Solution:

If you get an angle magnitude vector to start with, you have to break it into components.  If the 25.2o is with the x-axis, then it is the trig angle, and the initial x and y components are:

Vx = 43.2cos(25.2o) = 39.089 m/s

Vy = 43.2sin(25.2o)  = 18.394 m/s

We are also told that the wall is 130 m away, so this is the horizontal displacement s.

We can solve the horizontal for the time:

s = ut + 1/2at2

a = 0, so

s = ut , 130. m = (39.089 m/s)t

t = 3.3258 s

Now we find the final vertical velocity using v = u + at

v = 18.394 m/s + (-9.81 m/s/s)(3.3258 s)

v = -14.232 m/s (it is already going down)

If we draw this vector, it looks like this Where the horizontal is 39.089 m/s, and the vertical is -14.232 m/s.  The magnitude (speed) is the hypotenuse = sqrt((39.089 m/s)2 + (-14.232 m/s)2) = 41.6 m/s, (Be careful of squaring negative numbers with your TI graphing calculator)

and the angle indicated is tan-1(14.232/39.089) = 20.0o (below the horizontal) .:.

Questions 4 and 5 are to see if you can solve an arc trajectory problem, and solve for obscure stuff somewhere in the middle of the trajectory