Answer:
Explanation:
Given
Launch angle =u
Initial Speed is ![v_0](https://tex.z-dn.net/?f=v_0)
Horizontal acceleration is ![a_x=a](https://tex.z-dn.net/?f=a_x%3Da)
At maximum height velocity is zero therefore
![v_f=v_i-gt](https://tex.z-dn.net/?f=v_f%3Dv_i-gt)
![0=v_0\sin u-gt](https://tex.z-dn.net/?f=0%3Dv_0%5Csin%20u-gt)
![t=\frac{v_0\sin u}{g}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv_0%5Csin%20u%7D%7Bg%7D)
Total time of flight ![T=2t=\frac{2v_0\sin u}{g}](https://tex.z-dn.net/?f=T%3D2t%3D%5Cfrac%7B2v_0%5Csin%20u%7D%7Bg%7D)
During this time horizontal range is
![R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}](https://tex.z-dn.net/?f=R%3Dv_o%5Ccos%20u%5Ccdot%202t-%5Cfrac%7Ba%282t%29%5E2%7D%7B2%7D)
![R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B2v_0%5E2%5Csin%20u%5Ccos%20u%7D%7Bg%7D-%5Cfrac%7B2av_0%5E2%5Csin%20%5Eu%7D%7Bg%5E2%7D)
For maximum range ![\frac{\mathrm{d} R}{\mathrm{d} u}=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20R%7D%7B%5Cmathrm%7Bd%7D%20u%7D%3D0)
![\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20R%7D%7B%5Cmathrm%7Bd%7D%20u%7D%3D%5Cfrac%7B2v_0%5E2%5Ccos%202u%7D%7Bg%7D-%5Cfrac%7B4av_0%5E2%5Csin%20u%5Ccos%20u%7D%7Bg%5E2%7D)
![\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)
![\tan 2u=\frac{g}{a}](https://tex.z-dn.net/?f=%5Ctan%202u%3D%5Cfrac%7Bg%7D%7Ba%7D)
![u=\frac{1}{2}tan ^{-1}\frac{g}{a}](https://tex.z-dn.net/?f=u%3D%5Cfrac%7B1%7D%7B2%7Dtan%20%5E%7B-1%7D%5Cfrac%7Bg%7D%7Ba%7D)
(b)If a =10% g
![a=0.1g](https://tex.z-dn.net/?f=a%3D0.1g)
thus ![u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}](https://tex.z-dn.net/?f=u%3D%5Cfrac%7B1%7D%7B2%7Dtan%5E%7B-1%7D%5Cfrac%7Bg%7D%7B0.1g%7D)
![u=42.14^{\circ}](https://tex.z-dn.net/?f=u%3D42.14%5E%7B%5Ccirc%7D)
I think it's B because when you graph you do see the relationship between the dependent and independent variable
Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m