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:

Explanation:
here we know that car starts from rest and continue to accelerate till it will reach to maximum speed of 20 m/s
so we will have



so car will accelerate till t = 10 then it will move with uniform speed
so the distance moved by the car till it accelerates is given as



now it will move with uniform speed for next 36 s
so we have

so total distance moved by the car is given as



Answer:
factor that bug maximum KE change is 0.52284
Explanation:
given data
vertical distance = 6.5 cm
ripples decrease to = 4.7 cm
solution
We apply here formula for the KE of particle that executes the simple harmonic motion that is express as
KE = (0.5) × m × A² × ω² .................1
and kinetic energy is directly proportional to square of the amplitude.
so
.............2

= 0.52284
so factor that bug maximum KE change is 0.52284
Answer:c
Explanation: the speed of object a changes but b travels at constant speed