A hot air balloon has just lifted off and is rising at a constant rate of 2.0 m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 12 m/s. If the passenger is 2.5 m above her friend when the camera is tossed, how high is she above her friend when the camera reaches her?

Part b) Also, at the point (10 [cm], 5 [cm])

FINAL 4 SOLVING PHYS AClasswork for Physics C X nts Movement - X /forms/d/e/1FAlpQLSc-vuR6367rnrEWMQSorxTmuCz0zMsOHzru-hRxIMkvTL S= d/t SOLVE for "d" * O d = s/t d = st d = t/s Option 4 S=d/t. SOLVE for "t" * t= s d t=s/d t=s+d KO

Please explain the steps to find the answer, which equation you are using and why you are using it. Also a small reflection about the problem.  Thanks.

DATA: In this experiment, a tennis ball is released from a certain height. While the ball falls down, time of release (t;) and time when the ball touches the ground (tr) are recorded using a stopwatch and slow motion video. The difference At = tr-ti is the time interval during the fall. For the first experiment (available with video), t=17.31s and tr= 17.94s, yielding At = 17.94s - 17.31s = 0.63s.

part d What is the non-relativistic momentum of an electron moving at this speed (81 % of the speed of light), in kilogram meters per second?   Part (e)  When the electron is moving at 81 % of the speed of light, how many times greater is the relativistic momentum than the non-relativistic momentum?   Part (g)  What is the ratio of the relativistic momentum to the classical momentum of a rocketship moving at 99.2 % of the speed of light, in kg ⋅ m/s?

Solve the problem exactly like the example problem attached. The problem you're solving has a black background.   Use equation Vf^2=Vi^2+2ax Givens (right train) X= Want t= ? Vi= Vf= a= -1 m/s   Givens (left train) X= Want t= ? Vi= Vf= a= -0.5 m/s

Please explain the photo attached in detail