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>
(a) 
The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by
where
I is the intensity of the wave
c is the speed of light
In this problem,
and substituting 
, we find the radiation pressure
(b) 
Since we know the cross-sectional area of the laser beam:
starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:
And then, since we know the mass of the atom
we can find the acceleration, by using Newton's second law:
Answer:
Kinetic energy is given by:
K.E. = 0.5 m v²
Susan has mass, m = 25 kg
Velocity with which Susan moves is, v = 10 m/s
Hannah has mass, m' = 30 kg
Velocity with which Hannah moves is, v' = 8.5 m/s
<u>Kinetic energy of Susan:</u>
0.5 m v² = 0.5 × 25 kg × (10 m/s)² = 1250 J
<u>Kinetic energy of Hannah:</u>
0.5 m v'² = 0.5 × 30 kg × (8.5 m/s)² = 1083.75 J
Susan's kinetic energy is <u>1250 J </u>and Hannah's kinetic energy is <u>1083.75 J</u>.
Since kinetic energy is dependent on mass and square of speed. Thus, speed has a greater effect than mass. As it is evident from the above example. Susan has greater kinetic energy due to higher speed than Hannah.
Complete Question
Part of the question is shown on the first uploaded image
The rest of the question
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Answer:
The net force exerted on the third charge is 
Explanation:
From the question we are told that
The third charge is 
The position of the third charge is 
The first charge is 
The position of the first charge is 
The second charge is 
The position of the second charge is 
The distance between the first and the third charge is
The force exerted on the third charge by the first is
Where k is the coulomb's constant with a value 
substituting values
The distance between the second and the third charge is
The force exerted on the third charge by the first is mathematically evaluated as
substituting values
The net force is
substituting values
Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s.
