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PLEASE HELP! I will give brainliest and ten points!
A person throws a rock horizontally off a cliff ata a speed of 9 m/s from a point 64.8 m above the ground. How long is the rock in the air?
13.2 s
26.4 s
3.64 s(my guess)
5.14 s

A package falls from an airplane in level flight at constant speed. If air resistance can be neglected, how does the motion of the package look to the pilot.
The package appears to travel in the shape of a parabola.
The package appears to fall straight downward.(my guess)
The package appears to travel backward.
The package appears to fall backward and downward.

A paintball is shot horizontally in the positive x direction. At time Trianglet after the ball is shot, it is 4 cm to the right and 4cm below it’s starting point. At time 2trianglet, what is the position of the ball relative to it’s starting point. Ignore air resistance.
12 cm to the right and 16 cm below
8cm to the right and 8 cm below
12 cm to the right and 8 cm below
8 cm to the right and 16 cm below(my guess)
A baseball pitcher throws a fastball horizontally at a speed of 42.0 m/s. Ignoring air resistance, how far does the ball drop between the pitcher’s mound and home plate, 60 ft 6in away?
0.945 m
2.15 m
1.02 m
0.382 m
A rescue plane spots a person floating in a lifeboat 55 m directly below and releases an emergency kit with a parachute. The package descends with a constant vertical acceleration of 6.91 m/s. If the horizontal speed of the plane was 70.6 m/s, how far is the package from the lifeboat when it hits the waves?
225 m
1.27 km
323 m
282 m

PLEASE HELP I will give brainliest and ten points A person throws a rock horizontally off a cliff ata a speed of 9 ms from a point 648 m above the ground How lo class=

Respuesta :

korealeeanengele

I need help with this too! I also do Connexus

1)

Answer:

t = 3.64 s

Explanation:

As we know that the speed of the rock in vertical direction is ZERO

so here we will have

[tex]\Delta Y = \frac{1}{2}gt^2[/tex]

[tex]64.8 = \frac{1}{2}(9.8)t^2[/tex]

[tex]64.8 = 4.9 t^2[/tex]

[tex]t = 3.64 s[/tex]

2)

Answer:

The package appears to fall straight downwards

Explanation:

Since the package is dropped from the plane so the horizontal speed of the plane and package will be same

So here the package will always appear vertically down the plane

So here it will appear to move vertically down

3)

Answer:

8 cm to the right and 16 cm below

Explanation:

After one tringlet the position of the object is

[tex]x = 4 cm[/tex]

[tex]y = 4 cm[/tex]

Now we know that

[tex]x = v_x t[/tex]

[tex]y = \frac{1}{2}gt^2[/tex]

so here x coordinate is proportional to the time while y coordinate is square proportional to the time

so here in x direction it will move double distance in double time while in y direction its distance becomes 4 times

so we have

[tex]x = 8 cm[/tex]

[tex]y = 16 cm[/tex]

4)

Answer:

y = 0.945 m

Explanation:

distance of the home plate

[tex]d = 60 ft 6 in[/tex]

[tex]d = 60.5\times 0.3048m[/tex]

[tex]d = 18.44 m[/tex]

now the time taken by ball to reach the home plate

[tex]t = \frac{d}{v}[/tex]

[tex]t =\frac{18.44}{42} = 0.44 s[/tex]

Now the distance dropped by the ball in same time

[tex]y = \frac{1}{2}gt^2[/tex]

[tex]y = \frac{1}{2}(9.8)(0.44^2)[/tex]

[tex]y = 0.945 m[/tex]

5.

Answer:

d = 282 m

Explanation:

Time taken by the packet to reach the waves is given as

[tex]y = \frac{1}{2}gt^2[/tex]

[tex]55 = \frac{1}{2}(6.91)t^2[/tex]

[tex]t = 3.99 s[/tex]

now in the same time distance moved by the packet

[tex]d = vt[/tex]

[tex]d = 70.6 \times 3.99[/tex]

[tex]d = 282 m[/tex]

7.

Answer:

Between B to C

Explanation:

While moving downwards under gravity the attraction force of Earth will increase the speed

So its speed will increase during its return journey from B to C

8.

Answer:

y = 2040 m

Explanation:

given that

[tex]v_x = 50 m/s[/tex]

[tex]\Delta x = 1000 m[/tex]

now the time taken by the ball to reach the given distance is

[tex]t = \frac{d}{v}[/tex]

[tex]t = \frac{1000}{50} = 20 s[/tex]

now in the same time the vertical distance moved by it

[tex]y = v_y t - \frac{1}{2}gt^2[/tex]

[tex]y = 200(20) - \frac{1}{2}(9.8)(20^2)[/tex]

[tex]y = 2040 m[/tex]

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