Aprojectile is launched with an initial speed of 42 feet per second. it is projected at an angle of 50 degrees. how far does the projectile travel? how much farther does it travel if it is launched with an initial speed of 84 feet per second?

1) 53.95 ft ( approx )

2) 161.85 ft. ( approx )

Step-by-step explanation:

Given,

Initial velocity,

Also, the projected angle,

1) Thus, the distance travelled horizontally ( Range of the projectile motion ),

Where, g is the acceleration due to gravity that having the value 32.2 ft per second,

Hence, the projectile covers 53.95 feet ( approx )

2) If

Then the range would be,

Thus, the additional distance = 215.80 ft - 53.95 ft = 161.85 ft.

Hence, It will travel 161.85 ft extraa.

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

d i think

step-by-step explanation: