PHYS 1010 Lecture 10: Lecture 10 Notes

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maroonchinchilla495

6 Oct 2016

School

University of Virginia

Department

Physics

Course

PHYS 1010

Professor

Neu Christopher

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Document Summary

Without gravity, a tossed object follows a straight line path. With gravity, the same object tossed at an angle follows a curved path. A combination of a horizontal component and a vertical motion. Ball travels the same horizontal distance in equal times. Paths of a cannonball shot at an upward angle. Vertical distance that a stone falls is the same vertical distance it would have fallen if it had been dropped from rest and been falling for the same amount of time (5t^2) Ex) 1s = 5m, 2s = 20m, 3s = 45m. Same range is obtained from two different launching angles when the angles add up to 90 degrees. Object thrown at an angle of 60 degrees has the same range as if it were thrown at an angle of 30 degrees. The earth surface drops a vertical distance of 5 meters for every 8000 meters tangent to the surface.

Related Questions

MAC, Inc. (the company you work for) has decided to move forwardwith a feasibility
study for a product it wishes to manufacture and sell. This productis a particle launcher that should have the capability, for everyactivation, of launching two particles into the air at twodifferent times and at two different angles of elevation, in anautomated fashion so that they both land at the same distance fromthe launch point at the same time. For example, let P1 and P2 betwo particles, let T1 and T2 be two times, and let theta1 andtheta2 be two angles of elevation. Once activated, the launchershould be able to operate so that particle P1 is launched at timeT1 with an angle of elevation theta 1, and particle P2 islaunched
at time T2 with an angle of elevation theta2.
The launcher must be configured by a technician, prior to eachactivation, so that the
initial speed is set for the launcher, and the distance from thelauncher to where both
particles will land is set. For each activation of the launcher,the initial speed will be the
same for both particles. Once the technician has set the initialspeed and desired distance,
the launcher must calculate the angles of elevation to use and howmuch time to wait in
between particle launches. Then, when the technician activates thelauncher, both
particles will be launched at the appropriate angles of elevationand at the appropriate
times so that they both land at the configured distance from thelauncher at the same time.
The launcher, based on its calculations, will change the angle ofelevation to be ready for
the second particle launch. The launcher, however, must flash awarning for the
technician if the information the technician used to configure thelauncher does not result
in a two-particle launching solution.
Certain assumptions must be made. It is assumed that that launchercan always change
the angle of elevation to get ready for the second particle launchfaster than the time
delay calculated for the second particle launch. It is assumed thatthe units for speed are
meters per second, that the units for distance are meters, and thatgravity is 9.8 meters per
second squared. It is assumed that gravity is the only force actingon the motion of the
particles after launch, and that there is no wind to be considered,so that the particles land
at a point exactly in line with the direction the launcher ispointed.
The questions that must be answered and written up in anengineering report are as
follows:
1. Given the initial speed and desired landing distance for theparticles, find the
mathematical expressions that must be used to find 1, 2, and thetime delay
between particle launches.
2. Determine under which conditions the launcher will not be ableto perform the
operation requested, and will consequently flash a warning to thetechnician.
3. Each particle launched will follow a path of motion that isparabolic. Show that
the vertex of the parabola for each particleâs path occurs at thesame distance from
the launcher.
4. Along the path of each launched particle, there will be amaximum curvature.
Find that maximum curvature for each launched particle.

In the latest Indian Jones film, Indy is supposed to throw agrenade from his car, which is going 95.0 , to his enemy's car,which is going 125 . The enemy's car is 13.0 in front of the Indy'swhen he lets go of the grenade.

If Indy throws the grenade so its initial velocity relative to himis at an angle of 45 above the horizontal, what should themagnitude of the initial velocity be? The cars are both travelingin the same direction on a level road. You can ignore airresistance.

Hint: The grenade moves in projectile motion, and convert the twovelocities given in the problem from km/hr to m/s. Being anexcellent student of P2, Indy knows that horizontal range of thegrenade must equal the distance that the enemies car is ahead atthe time the grenade is thrown plus the distance the enemies cartravels while the grenade is in the air. This distance is given byv_rel*t, where v_rel the relative velocity of the enemies carrelative to the Indy's and t is the time in the air.

Solve for time that the grenade is in the air in terms ofv_0.

Use the range equation to get the grenade distance as a function ofv_0

Set R=(initial separation)+v_rel*t and get a 2nd order polynomialfor v_0. Use quadratic equation to get v_0. This is the magnitudeof the velocity vector relative to Indy.

and...
Find the magnitude of the velocity relative to the earth.

Hint: Use Galileo's equation where Frame A is ground, Frame B isIndy's car and the object of interest P is the granade.

Use your answer in part A for speed and given direction tocalculate the x and y components of the velocity with respect toIndy's car. The velocity of Indy's car with respect to the groundis straightforward to write in x and y components.

In the latest Indian Jones film, Indy is supposed to throw agrenade from his car, which is going 82.0 km/h, to his enemy's car,which is going 125km/h. The enemy's car is 14.6m in front of theIndy's when he lets go of the grenade.

=====>Part A: What is the Vo ? in km/h

If Indy throws the grenade so its initial velocity relative to himis at an angle of 45 ^\circ above the horizontal, what should themagnitude of the initial velocity be? The cars are both travelingin the same direction on a level road. You can ignore airresistance.

Hint: The grenade moves in projectile motion, and convert the twovelocities given in the problem from km/hr to m/s. Being anexcellent student of P2, Indy knows that horizontal range of thegrenade must equal the distance that the enemies car is ahead atthe time the grenade is thrown plus the distance the enemies cartravels while the grenade is in the air. This distance is given byv_rel*t, where v_rel the relative velocity of the enemies carrelative to the Indy's and t is the time in the air.

Solve for time that the grenade is in the air in terms ofv_0.

Use the range equation to get the grenade distance as a function ofv_0

Set R=(initial separation)+v_rel*t and get a 2nd order polynomialfor v_0. Use quadratic equation to get v_0. This is the magnitudeof the velocity vector relative to Indy.

=====>Part B: What is Vo in km/h?

Find the magnitude of the velocity relative to the earth.

Hint: Use Galileo's equation where Frame A is ground, Frame B isIndy's car and the object of interest P is the granade.

Use your answer in part A for speed and given direction tocalculate the x and y components of the velocity with respect toIndy's car. The velocity of Indy's car with respect to the groundis straightforward to write in x and y components.

PHYS 1010 Lecture 10: Lecture 10 Notes

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