Exam 1 practice problems solutions

Exam 1 practice problems solutions - Exam #1 practice...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Exam #1 practice problems solutions Physics 203 1. You are standing next to the Physics Lecture Hall. Your friend has climbed up unto the roof and is sitting 6.00 m above you with his legs dangling over the edge. His cell phone starts ringing (he left it with you before he climbed up). You toss him his cell but you toss it way too hard. It goes straight up passing your friend and reaches its apex (highest point) a full 3.00 m above his head. He catches it on the way down and answers the phone. It’s his mother, checking to make sure he’s not doing anything he shouldn’t be. a. Draw position, velocity, and acceleration graphs for the phone for the time interval that it was in flight. b. Predict how fast you tossed the phone (meaning, how fast it was traveling right after it left your hand). c. Predict how long the phone was in flight. d. What assumptions, if any, did you make in parts a c.? I’ll choose the origin of coordinates to be the position of the phone just as it leaves your hand, with the positive direction upward. This moment will also be the origin in time. To predict how fast I tossed the phone I’ll use the equation of kinematics that doesn’t involve time. The initial moment will be just as the phone leaves your hand and the final moment will be just as it reaches its apex.  22 2 2 0 2 9.8 9.00 0 13.3 fi if f i vv a xx vva x x m s m m s   0.5 1 1.5 2 t s 2 4 6 8 x m 1 2 t s -5 5 10 v m s 1 2 t s -20 -15 -10 a m s 2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
To predict the flight time I’ll use the equation that doesn’t involve the final velocity since I do not know what it is. The initial moment is when the phone leaves your hand and the final moment is when the phone is caught by your friend.  2 2 2 22 1 2 1 6.00 0 13.3 0 9.8 0 2 4.9 13.3 6.00 0 fi i f i f i ff xx v t t a t t mm s t m s t ms t mst m     This quadratic equation has two solutions, 0.57 s and 2.14 s . This is because the cell phone is at 6.00 m at two different moments, once on the way up, and then later when your friend catches it. So, it’s the larger time that represents the entire flight time. Since I’m using kinematics I’m assuming that the acceleration of the cell phone is constant. This will be the case only if air resistance is not affecting the motion of the cell phone. This is probably reasonable since the cell phone is never moving very quickly. 2. A secretive group known as The Dharma Initiative has built a hidden underwater base just off a remote island in the South Pacific. This base is called The Looking Glass and its purpose is to jam all transmissions coming to or from the island. The people working in The Looking Glass need supplies regularly, which are dropped by boat from above.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/22/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell.

Page1 / 20

Exam 1 practice problems solutions - Exam #1 practice...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online