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Phoenix [80]
2 years ago
9

Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.

On his first try, Robin looses the arrow at v0=35.0 m/sv0=35.0 m/s at an angle of θ=30.0°θ=30.0° above the horizontal. The arrow has an initial height of y0=1.50 m,y0=1.50 m, and its tip is x=60.0 mx=60.0 m away from the target orange. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has reached the horizontal position xx of the orange? Use g=9.81 m/s2g=9.81 m/s2 for the acceleration due to gravity.
Physics
1 answer:
tensa zangetsu [6.8K]2 years ago
4 0

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

<u>h' = 55.3 m</u>

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den301095 [7]

Answer:

whats the question

Explanation:

where your question

4 0
3 years ago
A 2 kg toy car moves at a speed of 5 m/s.<br> What is the kinetic energy of the car?
ANTONII [103]

Answer:

<h2>The answer is 25 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

KE =  \frac{1}{2} m {v}^{2}  \\

m is the mass

v is the velocity

From the question we have

KE =  \frac{1}{2}  \times 2 \times  {5}^{2}  \\  = 1 \times  {5}^{2}

We have the final answer as

<h3>25 J</h3>

Hope this helps you

5 0
2 years ago
The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.
Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0
3 years ago
Guys please helpp!!!!1
Setler79 [48]

Answer:

Position A/Position E

K = E, U = 0

Position B/Position D

K = (1-x)\cdot E, U = x\cdot E, for 0 < x < 1

Position C

K = 0, U = E

Explanation:

Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy (E) is the sum of gravitational potential energy (U) and translational kinetic energy (K), all measured in joules. In addition, gravitational potential energy is directly proportional to height (h) and translational kinetic energy is directly proportional to the square of velocity.

Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:

Position A/Position E

K = E, U = 0

Position B/Position D

K = (1-x)\cdot E, U = x\cdot E, for 0 < x < 1

Position C

K = 0, U = E

3 0
2 years ago
A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s
natima [27]

Answer:

The force of friction that acts on him is

F_k=567N

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

F=m*a

F=F_w-F_k

m*a=F_w-F_k

F_w=81kg*10m/s^2=810N

Sole to Fk

81kg*3m/s^2=810N-F_k

F_k=810N-243N

F_k=567N

4 0
2 years ago
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