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Yuri [45]
3 years ago
7

A student tosses a ball horizontally from a balcony to a friend 3.8 meters down below them. How long does the ball take to reach

their friend?
Physics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

The time it takes the ball to fall 3.8 meters to friend below is approximately 0.88 seconds

Explanation:

The height from which the student tosses the ball to a friend, h = 3.8 meters above the friend

The direction in which the student tosses the ball = The horizontal direction

Given that the ball is tossed in the horizontal direction, and not the vertical direction, the initial vertical component of the velocity of the ball = 0

The equation of the vertical motion of the ball can therefore, be represented by the free fall equation as follows;

h = 1/2 × g × t²

Where;

g = The acceleration due gravity of the ball = 9.81 m/s²

t = The time of motion to cover height, h

Then height is already given as h = 3.8 m

Substituting gives;

3.8 = 1/2 × 9.81 × t²

t² = 3.8/(1/2 × 9.81) ≈ 0.775 s²

∴ t = √0.775 ≈ 0.88 seconds

The time it takes the ball to fall 3.8 meters to friend below is t ≈ 0.88 seconds.

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A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no
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Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = A_{y} / A

    Aₓ = A cos 160 = 2 cos 160

    A_{y}  = A sin160 = 2 sin160

    Aₓ = -1,879 mi

    A_{y}  = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = B_{y} / B

    Bₓ = B cos 260

     B_{y}  = B sin 260

    Bₓ = 4 cos 260

     B_{y}  = 4 sin 260

     Bₓ = -0.6946mi

     B_{y}  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = C_{y} / C

     Cₓ = C cos 15

     C_{y} = C sin15

     Cₓ = 3 cos 15

    C_{y} = 3 sin 15

     Cₓ = 2,898 mi

    C_{y} = 0.7765 mi

Now we can find the final position of the person

    X = Aₓ + Bₓ + Cₓ

    X = -1.879 -0.6949 + 2.898

    X = 0.3241 mi

    Y = A_{y} +  B_{y} + C_{y}

    Y = 0.684 - 3.939 +0.7765

    Y = -2.4785 mi

a) We use Pythagoras' theorem

     R = √ (x2 + y2)

     R = √ (0.3241 2 + (-2.4785) 2)

     R = 2.4996 mi

     R = 2.5 mi

b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

     θ = tan⁻¹ (-7,647)

     θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

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