Physics, 13.09.2019 22:30 nickkungu92

1. a man stands on top of a 25 m tall cliff. the man throws a rock upwards at a 25-degree angle above the horizontal with an initial velocity of 7.2 m/s. what will be the vertical position of the rock at t = 2 seconds? what will be the horizontal position of the rock at that time?
2. a cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. if the cannon is on a mounted wagon 1.5 m tall, how far would the cannonball travel before it lands on the ground?
3. look at the following picture. what would be the resultant vector of a+b?
4. an airplane undergoes the following displacements: first, it flies 72 km in a direction 30° east of north. next, it flies 48 km due south. finally, it flies 100 km 30° north of west. using analytical methods, determine how far the airplane ends up from its starting point.
5.the following picture shows a golf ball being hit and given an initial velocity of v0. the ball is hit at an unknown angle above the ground. what are two values that are known, even in the absence of all other numbers?

1. a man stands on top of a 25 m tall cliff. the man throws a rock upwards at a 25-degree angle abov

Answers: 2

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Answers

Answer from: bstine6678

These are the answers for 1, 2 and 3

Explanation:

Sorry I couldn't help you with 4 and 5

1. a man stands on top of a 25 m tall cliff. the man throws a rock upwards at a 25-degree angle abov

Answer from: rainbowsadie7140

y = 11.48 m

x = 13.0 m

Explanation:

Components of initial velocity is given as

v_x = 7.2 cos25 = 6.52 m/s

v_y = 7.2 sin25 = 3.04 m/s

Now after t = 2 s the vertical position is given as

$y = y_0 + v_y t + \frac{1}{2}a_y t^2$

$y = 25 + 3.04(2) - \frac{1}{2}(9.8)(2^2)$

y = 11.48 m

Now horizontal position is given as

x = v_x t

$x = 6.52 \times 2$

x = 13.04 m

d = 26.6 m

Explanation:

Initial position on y axis is given as

y = 1.5 m

velocity of ball in y direction

v_y = 0

now we have

$\Delta y = v_y t + \frac{1}{2}gt^2$

$1.5 = \frac{1}{2}(9.8) t^2$

t = 0.55 s

now the distance moved by the ball in horizontal direction is given as

d = v_x t

$d = 48.1 \times 0.55$

d = 26.6 m

$A + B = 12.42\hat i + 0.35 \hat j$

Explanation:

Here we can see the two vectors inclined at different angles

so two components of vector A is given as

$A = 11.3 cos21\hat i - 11.3 sin21\hat j$

$A = 10.55 \hat i - 4.05 \hat j$

Similarly for other vector B we have

$B = 4.78cos67 \hat i + 4.78 sin67\hat j$

$B = 1.87\hat i + 4.4 \hat j$

now we need to find A + B

so we have

$A + B = (10.55\hat i - 4.05\hat j) + (1.87\hat i + 4.4 \hat j)$

$A + B = 12.42\hat i + 0.35 \hat j$

d = 81.86 m

Explanation:

Displacement of airplane is given as

d_1 = 72 km in direction 30 degree East of North

$d_1 = 72sin30\hat i + 72cos30\hat j$

$d_1 = 36\hat i + 62.35\hat j$

d_2 = 48 km South

$d_2 = -48\hat j$

d_3 = 100 km in direction 30 degree North of West

$d_3 = -100 cos30\hat i + 100 sin30\hat j$

$d_3 = -86.6\hat i + 50\hat j$

so net displacement is given as

d = d_1 + d_2 + d_3

$d = 36\hat i + 62.35\hat j - 48\hat j - 86.6\hat i + 50 \hat j$

$d = -50.6\hat i + 64.35\hat j$

now magnitude of displacement is given as

$d = \sqrt{50.6^2 + 64.35^2}$

d = 81.86 m

1) final speed at which it will hit the ground again

2) acceleration during the motion of the ball

Explanation:

As we know that the speed at which the ball is thrown is always same to the speed by which it will hit back on the ground

so we know that final speed will be same as initial speed

Also we know that during the motion the acceleration of ball is due to gravity so it will be

a = - g

Answer from: Quest

Is a vector quantity that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

Answer from: Quest

conservation of mass

mass before something happens = mass afterwards

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