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svlad2 [7]
2 years ago
9

Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How long will the bal

l be in the air before it lands? Round your answer to the nearest tenth. s How far will the football travel before it lands? Round your answer to the nearest tenth. m
Physics
2 answers:
RUDIKE [14]2 years ago
6 0

Answer:

1.2s for the first one

15.1m for the second one on time4learning

Explanation:

Step2247 [10]2 years ago
4 0

Answer:

T = 1.2 s

T = 15.1 m = 15 m

Explanation:

This is a case of projectile motion:

TOTAL TIME OF FLIGHT:

The formula for total time of flight in projectile motion is:

T = 2 V₀ Sinθ/g

where,

T = Total Time of Flight = ?

V₀ = Launch Speed = 13.9 m/s

θ = Launch Angle = 25°

g = 9.8 m/s²

Therefore,

T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)

<u>T = 1.2 s</u>

<u></u>

RANGE OF BALL:

The formula for range in projectile motion is:

R = V₀² Sin2θ/g

where,

R = Horizontal Distance Covered by ball = ?

Therefore,

T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)

<u>T = 15.1 m = 15 m</u>

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Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

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E= Potential energy

h = Vertical Height

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To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

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7 0
3 years ago
A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t
scoray [572]

Answer:

a)    F₁ = 267.3 N,   N₁ = 1300 N,  b)    μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

           

let's use subscript 1 for the ladder and 2 for the firefighter

            ∑ τ = 0

          -W₁ x₁ - W₂ x₂ + N₁ y = 0

           N₁ = \frac{W_1 x_1 + W_2 x_2}{y}          (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

         cos 60 = x₁ / d₁

         x₁ = d₁ cos 60

         x₁ = 7.5 cos 60

         x₁ = 3.75 m

for the firefighter d₂ = 4 m

         cos 60 = x₂ / d₂

         x₂ = d₂ cos 60

          x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

         sin 60 = y / d₃

         y = d₃ sin 60

         y = 15 sin 60

         y = 13 m

we look for the Normal by substituting in equation 1

         N₂ = \frac{500 \ 3.75 \ + 800 \ 2}{13}

         N₂ = 267.3 N

now let's use the translational equilibrium relations

 X axis

           F₁ - N₂ = 0

           F₁ = N₂

           F₁ = 267.3 N

Axis y

          N₁ - W₁ -W₂ = 0

          N₁ = W₁ + W₂

          N₁ = 500 + 800

          N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

          x₂ = 9 cos 60

          x₂ = 4.5 m

we substitute in 1

          N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}  

          N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

          fr = F₁ = N₂

          fr = 421.15 N

friction force has the expression

          fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

          μ = fr / N₁

          μ = 421.15 / 1300

          μ = 0.324

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Your answer is 2.5 x 10^(11).

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