Ask question

Ask question

All categories

3 years ago

8

A projectile is launched at an angle of 45 degrees with an initial speed of 32 feet per second. How far does the projectile trav

el?

1 answer:

Hatshy [7]3 years ago

6 0

Answer:

As the first part comes to the rest means, then it will fall under gravity directly.

First of all maximum height,

H=

g

u²sin²θ

H=

2×10

400×(

2

1

)

H=10m.

As initially the full projectile was supposed to land at A but due to explosion first part land at A', so to make the centre of mass lies at A the second part should lie at B.

f they don't break (explode) range OA

OA=

g

u²sin2θ

OA=400×sin45×

10

2

You might be interested in

Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

a.

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

b.

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

a.

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

b.

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

6 0

3 years ago

Write the square of the binomial as a polynomial: (-1 1/3p + 6q)^2

Thepotemich [5.8K]

Answer:

$\frac{16}{9}$ p² -16pq + 36q²

Step-by-step explanation:

Given

(- $\frac{4}{3}$ p + 6q)²

= (- $\frac{4}{3}$ p + 6q)(- $\frac{4}{3}$ p + 6q)

Each term in the second factor is multiplied by each term in the first factor, that is

- $\frac{4}{3}$ p(- $\frac{4}{3}$ p + 6q) + 6q(- $\frac{4}{3}$ p + 6q)

= $\frac{16}{9}$ p² - 8pq - 8pq + 36q² ← collect like terms

= $\frac{16}{9}$ p² - 16pq + 36q²

5 0

4 years ago

What is the equation of the line<br> with the points (1, 4) and (5, 12)?<br> 12

DerKrebs [107]

Answer:

y = 2x + 2

Hope you have a great day!

7 0

3 years ago

Simplify 3 x (7 - 2) + 6

Gnesinka [82]

The answer is 21.

if u look at (7-2) that gives you 5 then 3x5 gives you 15 and then you add 15+6 to get your answer of 21.

8 0

3 years ago

From a square sheet a 3 cm strap was cut off. The left rectangle has a surface of 70 cm^2. How long was the initial side of the

max2010maxim [7]

Side of square will be 10 cm

5 0

3 years ago