Homework Helper on projectile motion.

1) A ball is thrown horizontally at a speed of 20 m/s off a building that is 15 m high.
a) How long is the ball in the air?
b) How far from the base of the building does the ball land?
c) How fast is the ball moving just before it hits the ground?
d) At what angle does the ball hit the ground?

2) A ball is thrown at an angle of 40 degrees above the horizontal at an initial speed of 30 m/s.
a) How high does it rise above its release point?
b) How long is it in the air (assuming it lands at the same elevation from which it was thrown)?
c) How far away does it land?

KINEMATIC EQUATIONS:
avg. velocity = d/t
avg velocity = (v_i + v_f)/2
d = 1/2(v_i + v_f)t
v_f = v_i + at;
d = v_i(t) + 1/2 at²;

v_f² = v_i² + 2ad.

Question 1:

Follow the patented Byrne method for projectile motion problems:
a) Diagram the problem. The projectile is represented by a horizontal arrow to the right (at a 0º angle). The magnitude of the arrow (the initial velocity, vi) is 20.0 m/s.

b) Create a coordinate system: Here, there is no upward motion, so I will call up "negative" and down "positive", to the right will be "positive" and to the left "negative."
c) Create an x-/y- data table and fill in the known variables and solve for time on the y-side.
Use the equations Ax = Acos(theta), Ay = Asin(theta) to find the x- and y- components of the initial velocity.  Because this is a horizontal projectile, we should know that the initial velocity (20 m/s) is all in the x-direction, and nothing in the y-direction.  However, we can also show it mathematically:

d) We now solve for t in the y- direction. We can use d = v_i(t) + 1/2 at²; 
Substitute with units: 15 m = 0 + (1/2) 9.81 m/s^2(t²)
t = 1.75 s .  This is the time of flight.

e) While we're still on the y-side, we can find the final y-velocity at the instant before the projectile hits the ground.  Because we now know 4 of the 5 kinematic variables on the y-side, we can use any equation that has v_f in it.  I prefer to work with equations without squared terms (because I may forget to square something or get the square root), so I will choose:

v_f = v_i + at;

v_f = 0 + 9.81 m/s²(1.75 s) = 17.2 m/s

f) To find how far away it lands (an x-displacement), I go over to the x-side of my table.  I can bring the time from the y-side over to the x-side.  Our data table now looks like this:

I see that there is only one equation I can use on the x-side: d= v_i t (because the x-acceleration is zero). I substitute and solve:

d = v_i t = 20 m/s (1.75 s) = 35 m.

g) To find the speed and angle of impact of the ball as it hits the ground, I need to take the final velocity in the y-direction and add it to the final velocity in the x-direction.  This final step is exactly the same as what we did when we added vectors algebraically.  We now have the x- and y- components of the final velocity. (Remember, because the x-acceleration is zero, the inital x-velocity does not change, so it is also the final x-velocity.)  We can sketch them head-to-tail to remind ourselves how to find the actual final velocity in 2 dimensions:

h) We can see from the sketch that the final velocity of the ball is the hypoteneuse of a right triangle with sides of 20 m/s (the final x-velocity) and 17.2 m/s (the final y-velocity).  Using Pythagorean Theorem, the final velocity is:  v_f = √(20² + 17.2²) = 26.4 m/s.

i) The angle of impact can be found using the inverse tangent function:  θ = tan^-1 (17.2/20) = 40.7º below the horizontal.

Question 2:

2) A ball is thrown at an angle of 40 degrees above the horizontal at an initial speed of 30 m/s.
a) How high does it rise above its release point?
b) How long is it in the air (assuming it lands at the same elevation from which it was thrown)?
c) How far away does it land?

a) Diagram the problem. The projectile is represented by an arrow 40 degrees above the horizontal. The magnitude of the arrow (the initial velocity, vi) is 30.0 m/s.

b) Create a coordinate system: Here, there is upward motion, so I will call up "positive" and down "negative", to the right will be "positive" and to the left "negative."
c) Create an x-/y- data table and fill in the known variables and solve for time on the y-side.
Use the equations Ax = Acos(theta), Ay = Asin(theta) to find the x- and y- components of the initial velocity:

d) We need to know 3 out of 5 variables on the y-side, or 2 out of 3 variables on the x-side.  For some problems, we need to draw from the "hidden information" to get one or more variables.  In this problem, we can solve for the maximum height by setting the final y-velocity equal to zero and solving for the y-displacement:

Since we know vi, vf, and a and we want d, we can use:

v_f² = v_i² + 2ad.

Substituting with units and solving:

0 = (19.3 m/s)² + 2 (-9.81 m/s²)d

d = 19 m. 

This is the maximum height to which the ball rises.  Our data table now looks like this:

e) While we're still on the y-side, we can find the time to reach this height.  Because we now know 4 of the 5 kinematic variables on the y-side, we can use any equation that has t in it.  I prefer to work with equations without squared terms (because I may forget to square something or get the square root), so I will choose:

v_f = v_i + at;

0 = 19.3 m/s + -9.81 m/s²(t)

t = 1.97 s.  This is only the time for the ball to reach the highest point.  Because the ball lands at the same elevation from which it was fired, we need to double this time to get the time of flight.  So the ball is in the air for a total time of 3.94 s.

f) To find how far away it lands (an x-displacement), I go over to the x-side of my table.  I can bring the time of flight which I just learned from the y-side over to the x-side.  Our data table now looks like this:

I see that there is only one equation I can use on the x-side: d= v_i t (because the x-acceleration is zero). I substitute and solve:

d = v_i t = 23 m/s (3.94 s) = 90.6 m.
 

Create your own problems and check your answers using the Projectile Motion Applet.
Practice your motion graphs using Motion Graphs for Constant Acceleration.