The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
Answer:
Explanation:
Moment of inertia of a disc = 1/2 M R²
Since mass is same for both and radius are r and 2r, their moment of inertia can be in the ratio of 1: 4 . Let them be I and 4I . Angular speed are ω₀ and - ω₀ .
We shall apply law of conservation of angular momentum .
initial total angular momentum
I x ω₀ - 4I x ω₀ = - 3Iω₀
Let final common angular momentum be ω
total final angular momentum = ( I + 4I ) ω
Applying law of conservation of angular momentum
( I + 4I ) ω = - 3Iω₀
ω = - 3 / 5 ω₀ .
b )
Initial total rotational K E
= 1/2 I ω₀² + 1/2 4I ω₀²
= 1/2 x5I ω₀²
Final total rotational K E
= 1/2 ( I + 4I ) ( - 3 / 5 ω₀ )²
= 1/2 x 9 / 5 I ω₀²
= 9 / 10I ω₀²
change in rotational kinetic energy = 9 / 10I ω₀² - 1/2 x5I ω₀²
(9/10 - 5/2) xI ω₀²
=( .9 - 2.5 )I ω₀²
= - 1.6 I ω₀² Ans
To find out time, you put distance over speed. So you would have to put 150 over 50. You divide 150 by 50 and you would get 3. So your answer is 3 hours.
