Projectile Problems practice PDF

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Practice Problems for Exam 1...

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Projectile Problems Practice A projectile is an object that is propelled through the air by an initial push, and is afterwards subject only to the effects of gravity. (We will ignore the effect of air resistance in all of our projectile problems!) The height of a projectile is measured from the ground up to the top of its path. It is the maximum vertical distance the projectile reaches. It is labeled dy. The distance fallen is also referred to as dy, but it will have a negative value, due to its downward direction. If you always use df-di (change of position) to find dy then the vertical displacement, no matter whether it is actually the height or the distance fallen, the minus signs ALWAYS turn out right and clues you in as to the reference frame you are in. The final vertical velocity will be vyf = vi +gt, since its initial velocity changes by –10m/s for each second it falls. The range of a projectile is the maximum horizontal distance that it reaches from the point where it was fired. This is labeled dx. The range of a projectile is COMPLETELY UNAFFECTED by gravity. It is determined SOLELY from knowing the horizontal velocity, vx, (usually a given value) and the total time that the projectile has been in the air, ttotal. The formula for finding the range is dx= vxttotal. (And yes, dx can also be called the horizontal displacement!) Did you notice that the time up + the time down usually equals the total time? This is only true if the projectile returns to the same level. Time up can differ from time down if the projectile doesn’t return to the same level. Common Sense Warning: If the projectile is fired horizontally, it begins to fall right away, and in such cases the total time is the SAME as the time down. One interesting fact to know is that the same range can be reached in two ways if the projectile is fired at an angle: if two projection angles add up to 90 degrees, each one ends up producing the same range (this is not strictly true unless air resistance is neglected, though)! The trajectory of a projectile is the path through the air that it follows. The shape of this path is a parabola. We know that a quadratic relationship results in a parabola whenever y=kx2. (where k is a proportionality constant, like the slope m would be in y = mx +b). This relationship for projectiles means that the distance up or down is proportional to the square of the time up or down. Galileo noticed this relationship while investigating balls rolling down inclined planes. He wrote the relationship as d = ½ at2. Of course, the acceleration here is the acceleration due to gravity, which is always constant at –9.8m/s2. Round this to g = -10m/s2, and now you have the formula for height as dy= ½ gt2. [Notice that if you are measuring the distance fallen (down), dy should be negative, and that if you are measuring the distance a projectile rises (up), dy should be positive. The t in this formula is the time up or the time down, not the total time. ] By the way, the full expression for finding the distance travelled either vertically or horizontally is this: df-di=Vit + 1/2at2. However, our special starting conditions reduce this to the simpler formulas shown above. So overall, we have: dx= Vxttotal (horizontally) and dy= 1/2gt2 (vertically), with g = -10 m/s2.

Summary of formulas for projectiles:

Horizontal motion

Vertical motion

Displacement: Δdx=vxt (How far?) (a.k.a., range )

Δdy = vit + 1/2 at2, (a.k.a.,distance fallen or height) reduces to df = ½ gt2, if vi = 0 and di = 0.

Time of fall: (How long?)

t2 = 2Δdy/g (a.k.a., falling time or tdown)

same as vertical part.

ttotal is tup + tdown Velocity: (How fast?)

vx is constant

Acceleration:

none in x direction

vy = vi + gt, reduces to vy = -10t, if vi = 0

ay = g = -10 m/s2

Remember WHY these motions occur: No net force is exerted on the projectile horizontally, so there is no horizontal acceleration and it continues its state of motion with no change of horizontal velocity (all a consequence of Newton’s 1st Law of Motion). However, since a net force does exist vertically (our friend gravity), there is a vertical acceleration (a constant downward one) and so the projectile’s vertical velocity increases by –10 m/s for each second of fall (all a consequence of Newton’s 2nd Law of Motion and the definition of acceleration as a change of velocity over time).

Projectile Problems to practice : (Use g = -10 m/s2 and the formulas above. SHOW ALL WORK--NOT JUST ANSWERS!) 1) Alice throws a beanbag horizontally out of a 4th floor window at 10 m/s. a) Find how long it takes the beanbag to hit the ground if it starts out 20 m above the ground. (Use Δdy= 1/2gt2) [solution: t2 = 2dy/g = 2(-20m)/-10m/s2 = +4 s2, so t = 2 sec]

b) What vertical speed will the bag have just as it hits the ground? (Use vf = vi +gt) [solution: vf = vi +gt = 0 + -10m/s2 (2s), so vf = -20 m/s and speed = 20m/s] c) How far will the beanbag travel, horizontally? (Use dx = vxt) [solution: dx = vxt = 10m/s(2s) = 20m] 2) A cannonball is fired horizontally at 100 m/s from the top of a castle wall that is 40 m high. a) Find how long before it reaches the ground level. [solution: t2 = 2dy/g = 2(-40m)/-10m/s2 = +8 s2, so t = 2.8 sec] b) Find the vertical speed that it has as it hits the ground. [solution: vf = vi +gt = 0 + -10m/s2 (2.8s), so vf = -28m/s and speed =28m/s]

c) Find the range of the cannonball. [solution: dx = vxt = 100m/s(2.8s) = 280m]

3) A projectile is fired at a 35° angle. What other projection angle will give the same range? ____55°____ [because any 2 projection angles with the same range must add up to 90°]

Projectile—Horizontal Distance 54 The pictures below depict cannonballs of two different masses projected upward and forward. The cannonballs are projected at various angles above the horizontal, but all are projected with the same horizontal component of velocity. Rank according to the horizontal distance traveled by the balls. A

B

C v = 23.1 m/s

v = 23.1 m/s 1 kg

30°

1 kg

30°

2 kg

v x = 20 m/s

v = 40 m/s

45°

v x = 20 m/s

v x = 20 m/s

E

D

v = 28.3 m/s

F

v = 40 m/s

v = 28.3 m/s

2 kg

45°

1 kg

v x = 20 m/s

60°

v x = 20 m/s

2 kg

60°

v x = 20 m/s

Largest 1________ 2________ 3________ 4________ 5________ 6_______ Smallest All distances traveled are the same. ____________ Please carefully explain your reasoning.

[Solution: since all have the same horizontal velocity, the longest range will depend on which is in the air the longest, which depends on vertical velocity. E and F are tied for the largest vertical velocities, so they will be in the air the longest and have the longest range. C and D are tied for the next highest vertical velocity, so they will tie for the 2nd longest range. A and B are tied for the lowest vertical velocities, so they are in the air the shortest time and tie for the shortest range. These conclusions depend on the horizontal velocity being the same in all cases. ]

Projectile—Time in Air 55 The pictures below depict cannonballs of two different masses projected upward and forward. The cannonballs are projected at various angles above the horizontal, but all are projected with the same vertical component of velocity. Rank according to the time the balls are in the air. A

B

C v y = 20 m/s

v y = 20 m/s

v y = 20 m/s

v = 23.1 m/s

v = 28.3 m/s

v = 40 m/s

45°

30°

1 kg

1 kg

1 kg

D

F

E

v = 23.1 m/s

v = 28.3 m/s

v = 40 m/s v y = 20 m/s

v y = 20 m/s

v y = 20 m/s 45°

30°

2 kg

60°

60°

2 kg

2 kg

Largest 1________ 2________ 3________ 4________ 5________ 6_______ Smallest All times are the same. ____________ Please carefully explain your reasoning.

How sure were you of your ranking? (circle one) Basically Guessed Sure 1 2 3 4 5 6 55

7

8

Very Sure 9 10

S. Heath

Physics Ranking Tasks

58

Mechanics

[Solution: since time in the air depends on vertical velocity, and all have the same vertical velocity, all will tie for the same time in the air.]...