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>
D-a is the answer can I get braillist
Answer:
Momentum, p = 5 kg-m/s
Explanation:
The magnitude of the momentum of an object is the product of its mass m and speed v i.e.
p = m v
Mass, m = 3 kg
Velocity, v = 1.5 m/s
So, momentum of this object is given by :
p = 4.5 kg-m/s
or
p = 5 kg-m/s
So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.
Answer:
Wt = 26.84 [N]
Explanation:
In order to solve this problem we must use the definition of work in physics. Which tells us that this is equal to the product of force by distance.
In this case, we must sum the works of the force applied by the box and the friction force that also acts on the box.
The friction force is defined as the product of the normal force by the coefficient of friction.
f = N*μ
where:
N = normal force = m*g [N] (units of Newtons)
m = mass = 72 [kg]
g = gravity acceleration = 9.81 [m/s²]
f = friction force [N]
μ = friction coefficient = 0.21
f = 72*9.81*0.21
f = 148.32 [N]
Now the total work:
Wt = WF - Wf
where:
Wt = total work [J] (units of Joules)
WF = work by the pushing force [J]
Wf = work done by the friction force [J]
Wt = (160*2.3) - (148.32*2.3)
Wt = 26.84 [N]
Note: The friction force exerts a negative work, because this force is acting in opposite direction to the movement, therefore the negative sign.
Explanation:
heat caoacity and heat is difference
