Answer:
distance=6.11m
Explanation:
A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How far does the player run before he catches the ball? Ignore air resistance. I got stuck because I wasn't sure which formula to use when approaching this problem. Does it involve an angle at all?
first of all we get the time it takes to reach the maximum height
then twice of the time it takes to reach maximum height will be the time of flight
from newtons equation of motion
v=u+at
v=0
u=6m/s
0=6-9.81t
t=.61s
the time of flight will be 1.22secs
how far it travels will then be d, the basketball player moves with a horizontal speed 2.5m/s towards the ball
distance=speed*time
distance=2.5m/s*1.22
distance=6.11m
The energy of a photon is given by:

where h is the Planck constant and f is the photon frequency.
We know the energy of the photon,

, so we can rearrange the equation to calculate the frequency of the photon:

And now we can use the following relationship between frequency f, wavelength

and speed of light c to find the wavelength of the photon:
ANY orbiting object travels fastest when it's closest to the central body.
Question is missing. Found on google:
<em>"Part A What is the acceleration of the ball? Express your answer to two significant figures and include the appropriate units. </em>
<em>Part B
</em>
<em>What is the net force on the ball during the hit? </em>
<em>Express your answer to two significant figures and include the appropriate units."</em>
Solution:
A) 
The acceleration of the ball is given by

where
v = 12 m/s is the final velocity
u = 0 is the initial velocity (the ball is stationary)
t = 2.0 ms = 0.002 s is the time of contact
Substituting,

B) 
The force on the ball can be found by using Newton's second law:

where
m = 140 g = 0.14 kg is the mass of the ball
is the acceleration
Substituting,
