An archer fires an arrow straight up into the air with a speed vo = 42 m/s. neglect air resistance. 1. find the maximum height h reached by the arrow, in meters. 2.write an expression for the time the arrow is in the air until it returns to launch height in terms of known quantities.

As per the question the arrow was projected into space with an initial velocity of 42 m/s

Hence

We have to calculate the maximum height achieved by the arrow and total time taken to come the same point at which it was projected up into the air

Here air resistance is neglected

When the arrow reaches it s maximum height it's velocity becomes zero.

Hence final velocity

Putting the equation of kinematics we get-

                  ⇒

                  ⇒    [here g is opposite to displacement]

                  ⇒1764=19.6×h

                  ⇒h

                  ⇒h=90 m

Now we have to calculate the total time of flight.

Now we have to calculate the time of ascent.

From the kinematics we know that

                                                             ⇒0=u+[-g]×t

                                                              ⇒

we know that time of ascent is equal to time of descent.

hence total time of flight=2×[u/g]

The total time of flight

                                            =

                                           = 8.571 second.

c)as speed changes, wavelength and frequency change.

everything changes!

compass can be used day and night, cloudy or sunny