Question

PLEASE HELP! I will give brainliest and ten points!

Answer 1

I need help with this too! I also do Connexus

Answer 2

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

$\Delta Y = \frac{1}{2}gt^2$

$64.8 = \frac{1}{2}(9.8)t^2$

$64.8 = 4.9 t^2$

$t = 3.64 s$

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

$x = 4 cm$

$y = 4 cm$

Now we know that

$x = v_x t$

$y = \frac{1}{2}gt^2$

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

$x = 8 cm$

$y = 16 cm$

4)

Answer:

y = 0.945 m

Explanation:

distance of the home plate

$d = 60 ft 6 in$

$d = 60.5\times 0.3048m$

$d = 18.44 m$

now the time taken by ball to reach the home plate

$t = \frac{d}{v}$

$t =\frac{18.44}{42} = 0.44 s$

Now the distance dropped by the ball in same time

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.8)(0.44^2)$

$y = 0.945 m$

5.

Answer:

d = 282 m

Explanation:

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

$y = \frac{1}{2}gt^2$

$55 = \frac{1}{2}(6.91)t^2$

$t = 3.99 s$

now in the same time distance moved by the packet

$d = vt$

$d = 70.6 \times 3.99$

$d = 282 m$

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

$v_x = 50 m/s$

$\Delta x = 1000 m$

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

$t = \frac{d}{v}$

$t = \frac{1000}{50} = 20 s$

now in the same time the vertical distance moved by it

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

$y = 200(20) - \frac{1}{2}(9.8)(20^2)$

$y = 2040 m$