Amodel rocket rises with constant acceleration to a height of 3.1 m, at which point its speed is 28.0 m/s. a. how much time does it take for the rocket to reach this height?
b. what was the magnitude of the rocket's acceleration?
c. find the height of the rocket 0.10 s after launch.
d. find the speed of the rocket 0.10 s after launch.

Explanation:

It is given that,

Height, h = 3.1 m

Initial speed of the rocket, u = 0

Final speed of the rocket, v = 28 m/s

(b) Let a is the acceleration of the rocket. Using the formula as :

(a) Let t is the time taken to reach by the rocket to reach to a height of h. So,

t = 0.22 seconds

(c) At t = 0.1 seconds, height of the rocket is given by :

h = 0.63 meters

(d) Let v' is the speed of the rocket 0.10 s after launch.

So,

v' = 12.64 m/s

Hence, this is the required solution.

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m