100 as momentum=mass×velocity
Answer:
Explanation:
Given
Launch angle =u
Initial Speed is 
Horizontal acceleration is 
At maximum height velocity is zero therefore



Total time of flight 
During this time horizontal range is


For maximum range 

![\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20R%7D%7B%5Cmathrm%7Bd%7D%20u%7D%3D%5Cfrac%7B2v_0%5E2%7D%7Bg%7D%5Cleft%20%5B%20%5Ccos%202u-%5Cfrac%7Ba%7D%7Bg%7D%5Csin%202u%5Cright%20%5D%3D0)


(b)If a =10% g

thus 

Answer:
f = 3.09 Hz
Explanation:
This is a simple harmonic motion exercise where the angular velocity is
w² =
to find the constant (k) of the spring, we use Hooke's law with the initial data
F = - kx
where the force is the weight of the body that is hanging
F = W = m g
we substitute
m g = - k x
k =
we calculate
k =
k = 3.769 10² m
we substitute in the first equation
w² =
w = 19.415 rad / s
angular velocity and frequency are related
w = 2πf
f =
f = 19.415 / 2pi
f = 3.09 Hz
Answer:
I would love to help but I don't know I'm so sorry
Answerana alyom kint gahda amshi
Explanation: