An Object moving at a velocity of 30 m/s slows to a stop in 7 seconds. What was its acceleration

The moment the body impacts the ground, two types of Forces are produced: the gravitational pull and the Normal Force. This aligns with Newton's third law, indicating that every action has an equal and opposite reaction. If the downward force of gravity is directed toward the earth, the reactionary force from the block acts upwards, equivalent to its weight:

F = mg

Where,

m = mass

g = gravitational acceleration

F = 5*9.8

F = 49N

Consequently, the answer is E.

The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height. 

<span>In this scenario, you can demonstrate it as follows: </span>

<span>the initial velocity is Vo </span>
<span>the launch angle is α </span>

<span>the initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity becomes </span>
<span>Vh = Vo×cos(α) </span>

<span>the total flight duration is the period required to return to a height of 0 m, thus </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 = gravitational acceleration = g (= -9.8 m/s²) </span>
<span>therefore </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 let's examine the horizontal distance. </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>therefore </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 points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set 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>As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>

Answer:

(a). The distance traveled is 7.06 km towards the west.

(b). Their displacement magnitude is 7.51 km.

Explanation:

The information given states that,

The geese initially move 4.0 km directly west, then alter course to the north at a 40° angle, covering an additional 4.0 km.

Based on the diagram,

(a). To determine the distance

We will apply the distance formula

Insert the values into the equation

The resultant distance is 7.06 km westward.

(b). We will find the total displacement's magnitude

Using the displacement formula

Insert the values into the equation

The total displacement magnitude is 7.51 km.

In conclusion, (a). The traveled distance is 7.06 km towards the west.

(b). The magnitude of their total displacement is 7.51 km.