Physics Help! Projectile motion with only Y information given - 'liquid in syringe' problem?

Question

Our class is having difficulty with this projectile motion problem that we have to have submitted by tomorrow afternoon. The problem is...

"When a nurse squeezes a syringe, the liquid squirts 3.43 cm into the air. With what speed does the liquid emerge from the syringe?"

She gave us multiple choices to pick from, and in case they are helpful, I'll include them as well.

a.) 0.410 m/s
b.) 0.820 m/s
c.) 1.64 m/s
d.) 2.28 m/s

Thanks for any help!
Phys Class in MN

Thanks for the help! I have solved the problem.

Answer 1

here we go kinetic energy at point of shot equals potencial energy at top of the squirt

1/2 m v^2 = mgh

m's cancel (Rained in seattle)

v = Square root (2gh)

check my algebra... Note, you must change cm to meters before you can answer

Answer 2

a million. First we want to discover how lengthy the cannonball is contained in the air, its "draw close time" in case you'll. This span of time is the time it takes for the cannonball to bypass up, and fall lower backtrack. that is thoroughly depending on the y-area of the speed, so we could use some trigonometry to discover this. in case you drew a suitable triangle, you'll discover that the y-comp onent: speed at y axis = (preliminary speed)sin(seventy 9) = 993sin79 = 974.756 m/s. 2. Now that you've the speed on the y axis, we may be able to now discover the time it takes for the cannon to land. we may be able to apply the equation very last speed = preliminary speed + (Acceleration)(Time) or Vf = Vi + at note that your aceeleration = -g, as a results of the indisputable fact that's pulling it downward so: Vf = Vi - gt. placing apart t: Vf + gt = Vi gt = Vi - Vf t = (Vi - Vf)/g all of us understand the preliminary p.c. contained in the y axis, that is 974.756 m/s. yet what about the perfect speed? that is a few thing interesting: once you throw a ball up at say a particular p.c. 2m/s, even as it falls back on your hand (assuming you kept your hand on a similar position) it is going to go back on your hand at a p.c. of -2m/s, the alternative of the preliminary speed. So contained in terms of the cannonball, i will assert that the perfect speed is -974.756 m/s. i understand my Vi, Vf, and g, so i will now come to a call for time. t = (974.756 - (-974.756))/9.8 = (974.756 + 974.756)/9.8 = 1949.512/9.8 = 198.9298 seconds. 3. Why do i want to understand the draw close time? it is because the ball is travelling on the x course for this a lot time, and after that factor, it stops shifting (except it keeps on rolling, yet that is yet another difficulty! enable's assume it continues to be positioned the position it lands.) besides, i will apply the equation p.c. = distance/time or distance = (p.c.)(time) = vt I have my time, yet i want my speed. remember we are searching for the area on the x-course, not the y, so we for the speed right here, we want to get the x-area of the projectile's speed. utilizing trigonometry back: speed at x axis = (preliminary speed)cos(seventy 9) = 993cos79 = 993(0.19) = 189.473 m/s. So, utilizing this on the previous equation, and the air-time: distance traveled alongside the horizontal course = vt = (189.473)(198.9298) = 37,691.826 meters.

Answer 3

E = mgh (potential energy)
E = 1/2mv² (kinetic energy)

The kinetic energy when it leaves the syringe all becomes potential energy at the top of the arc, so...

mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = sqrt(2gh)

v = sqrt (2 x 9.8m/s² x .0343m)
v = sqrt (0.673 m²/s²)
v = 0.820 m/s (b)

Answer 4

kinetic energy equals potential energy.

If your entire class is having a problem, your instructor needs to do more work with you.