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oee [108]
1 year ago
5

A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it

reach the ground? How far from the point of projection would it strike?
Physics
1 answer:
Crazy boy [7]1 year ago
4 0

Answer:

20.41 s

3534.80 m

Explanation:

<em><u>In how many seconds will it reach the ground?</u></em>

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

  • 200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

  • v = v₀ + at

Substitute known values into the equation.

  • 0 = 200 · sin(30) + (-9.8)t
  • -200 · sin(30) = -9.8t
  • t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

  • 2t = 20.40816327

The body will reach the ground in 20.41 seconds.

<em><u>How far from the point of projection would it strike? </u></em>

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

  • Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

  • Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
  • Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

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What is the instantaneous velocity of a freely falling object 9.0 s after it is released from a position of rest? Express your a
Umnica [9.8K]

Answer:

So instantaneous velocity after 9 sec will be 88.2 m/sec              

Explanation:

We have given time t = 9 sec

As the object is released from rest so its initial velocity u = 0 m/sec

We have to find its final velocity v

Acceleration due to gravity g=9.8m/sec^2

From first equation of motion we know that v=u+gt

v=0+9.8\times 9=88.2m/sec

So instantaneous velocity after 9 sec will be 88.2 m/sec  

5 0
2 years ago
You are piloting a small airplane in which you want to reach a destination that is 750 km due north of your starting location. O
alexira [117]

Answer:

v_wind = 101.46 km / h   ,  θ = 61.8

Explanation:

This is a velocity composition exercise.

Let's do the problem in parts. Let's start by knowing the speed of the plane without air.

           v = d / t

           v = 750 / 3.14

           v = 238.85 km / h

This is the speed of the plane relative to the Earth and it does not change.

In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and v_{y}

Let's use trigonometry to find the components of the plane's speed

          cos θ = v_N / v

          sin θ  = v_W / v

          v_N = v cos θ

          v_W = v sin θ

           

let's calculate

          V _N = 238.85 cos 22 = 221.46 km / h

           v_W = -238.85 sin 22 = -89.47

the negative sign is because the plane is going west and the positive sign is the east direction.

As it indicates that the destination of the avine is towards the north, the x component of the wind must be

              vₓ - v_W = 0

              vₓ = v-w

              vₓ = 89.47 km / h

in the direction to the East.

Now let's analyze the component of the wind in the Nort-South direction,

Indicate the travel time, let's calculate the speed that the component must have the speed of the plane

             v_total = d / t

             v_total = 750 / 4.32

             v_total = 173.61 km / h

This is the final speed of the plane, which can be written

              v_total = v_n - vy

               vy = v_n - v_total

               vy = 221.46 - 173.61

               vy = 47.85 km

this component is directed towards the south

Let's use the Pythagorean Theorem, to find the magnitude

             v_wind² = vₓ² + vy²

             v_wind = √ (89.47² + 47.85²)

             v_wind = 101.46 km / h

the address will then be found using trigonometry

             θ = Vy / vx

             θ = tan⁻¹ (vy / vx)

             θ = tan⁻¹1 (47.85 / 89.47)

             θ = 28.14

Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value

                  90- θtea = 90- 28.2

                  θ = 61.8

East of South

5 0
1 year ago
(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0
2 years ago
A uniform cube with mass 0.700 kg and volume 0.0270 m3 is sitting on the floor. A uniform sphere with radius 0.400 m and mass 0.
Sav [38]

Answer:

  44 1/3 cm

Explanation:

The cube has an edge length of ∛0.027 m = 0.3 m, so a center of mass (CoM) 15 cm above the floor.

The sphere's center of mass is 40 cm above the top of the cube, so is 70 cm above the floor. The weighted average of the CoM locations is ...

  ((15 cm)(0.700 kg) +(70 cm)(0.800 kg))/(0.700 kg +0.800 kg)

  = (10.5 kg·cm +56 kg·cm)/(1.500 kg) = 44.333... cm

The center of mass of the two-object system is 44 1/3 cm above the floor.

_____

<em>Comment on the units</em>

We're not familiar with "hcm" as a unit. We presume that you can convert the given answer to the units you desire.

6 0
2 years ago
Two particles collide and stick together. If no external forces act on the two particles, which of the following is correct for
Semmy [17]

Answer:

c.\Delta p=0 and \Delta k

Explanation:

We are given that two particles collide  and stick together.

If there is no external force act on the two particles then ,it is inelastic collision.

Inelastic collision: There is some loss of kinetic energy but the momentum is conserved.

According to law of conservation of momentum

Initial momentum=Final momentum

Change in momentum=Final momentum-Initial momentum=0

Change in momentum=\Delta p=0

Initial kinetic energy is greater than final kinetic energy.

Change in kinetic energy=Final kinetic energy-kinetic energy=- negative

\Delta k

Hence, option c is true.

c.\Delta p=0 and \Delta k

4 0
2 years ago
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