Help with these below and the above. I will give brainliest and 20 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
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.
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

A baseball pitcher throws a football 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 6 in away? Note: 1ft = 0.03048 m.
0.945
2.15
1.02
0.382

Since the package is dropped from plane so the package and plane both will have same speed in horizontal direction so with respect to plane the package will always remain in same horizontal position.

Now here since the package is also moving downwards due to free fall so it will always seen as dropped down while we are watching it from the plane

The package appears to fall straight downward.

#3

Let the time trianglet is "t"

now in this interval of time if it falls down by 4 cm

[tex]h = v_y *t + 0.5 gt^2[/tex]

[tex]4 = 0 + 0.5* 9.8 * t^2[/tex]

[tex]t^2 = \frac{4}{4.9}[/tex]

now in same time it covers horizontal distance

[tex]d = v_x * t[/tex]

[tex]4 = \sqrt{\frac{4}{4.9}} * v_x[/tex]

[tex]v_x = 4.43 m/s[/tex]

now after two trianglet the horizontal position will be

[tex]x = v_x * t[/tex]

[tex]x = 4.43 * 2*\sqrt{\frac{4}{4.9}}[/tex]

[tex] x = 8 cm[/tex]

[tex]y = v_y * t + 0.5 g t^2[/tex]

[tex]y = 0 + 0.5 * 9.8 * 4 * \frac{4}{4.9} = 16 cm[/tex]

so it is 8 cm to the right and 16 cm below

#3

Speed horizontal = 42 m/s

distance covered = 60 ft 6 inch

[tex]d = 60.5 * 0.3048 = 18.44[/tex]

time taken to cover the distance

[tex]x = v* t[/tex]

[tex]18.44 = 42 * t[/tex]

[tex]t = 0.44 s[/tex]

now the distance by which it is dropped

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

[tex]y = 0.5 *9.8*0.44^2 = 0.944 m[/tex]

#4

here the velocity of projectile will increase when it will fall under gravity

So here it we can see initially it is moving up against the gravity so its speed will increase but after that its speed will increase when it starts falling down

So the correct answer is Between B to C

#5

when x coordinate is 1000

time taken to reach that point will be

[tex]x = v_x * t[/tex]

[tex]1000 = 50 * t[/tex]

[tex]t = 20 s[/tex]

now in y direction we have

[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]