Answer:
We are given the trajectory of a projectile:
y=H+xtan(θ)−g2u2x2(1+tan2(θ)),
where H is the initial height, g is the (positive) gravitational constant and u is the initial speed. Since we are looking for the maximum range we set y=0 (i.e. the projectile is on the ground). If we let L=u2/g, then
H+xtan(θ)−12Lx2(1+tan2(θ))=0
Differentiate both sides with respect to θ.
dxdθtan(θ)+xsec2(θ)−[1Lxdxdθ(1+tan2(θ))+12Lx2(2tan(θ)sec2(θ))]=0
Solving for dxdθ yields
dxdθ=xsec2(θ)[xLtan(θ)−1]tan(θ)−xL(1+tan2(θ))
This derivative is 0 when tan(θ)=Lx and hence this corresponds to a critical number θ for the range of the projectile. We should now show that the x value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace tan(θ) with Lx in the second equation from the top and solve for x.
H+L−12Lx2−L2=0.
This leads immediately to x=L2+2LH−−−−−−−−√. The angle θ can now be found easily.
Answer:The wavelength of a 1-kHz sound traveling in water would be 150 cm.
Explanation:This is because in the equation, v(velocity) = 1500 m/s, and f (frequency) = 1 kHz.
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
Answer:
Angle Of Incidence Formula The angle of incidence is equal to the reflected angle through the law of reflection. The angle of incidence and the angle of reflection is always equal, and they are both on the same plane along with the normal. 2,44,451
Explanation:
that could help you with work.
Time in the universe vs the population