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:
Its approx location is (5.18,1.9)
Explanation:
Using F( 5,2) = ( xy-1, y²-11)
= ( 5*2-¹, 2²-11)
= (9,-5)
= so at point t=1.02
(5,2)+(1.02-1)*(9,-5)
(5,2)+( 0.02)*(9,-5)
(5+0.18, 2-0.1)
= ( 5.18, 1.9)
Answer:
B. Only wavelengths between infrared and ultraviolet
Explanation:
That is the visible spectrum if you look up a diagram of the light spectrum.
<span>95 km/h = 26.39 m/s (95000m/3600 secs)
55 km/h = 15.28 m/s (55000m/3600 secs)
75 revolutions = 75 x 2pi = 471.23 radians
radius = 0.80/2 = 0.40m
v/r = omega (rad/s)
26.39/0.40 = 65.97 rad/s
15.28/0.40 = 38.20 rad/s
s/((vi + vf)/2) = t
471.23 /((65.97 + 38.20)/2) = 9.04 secs
(vf - vi)/t = a
(38.20 - 65.97)/9.04 = -3.0719
The angular acceleration of the tires = -3.0719 rad/s^2
Time is required for it to stop
(0 - 38.20)/ -3.0719 = 12.43 secs
How far does it go?
65.97 - 38.20 = 27.77 M</span>