MATHEMATICS COLLEGE

An attacker at the base of a castle wall 4 m high throws a rock straight up with speed 7.5 m/s from a height of 1.5 mabove the ground.
(a) Will the rock exceed the top of the wall?
(b) If so, what is its speed when it reaches the top of the wall?
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(c) If we don’t want the stone to exceed the top of the wall (we want the top of the wall to be the maximum height)
what the initial speed that the stone must have?

Answers

Answer 1

Answer:

Answer:

a) Yes, b) $v\approx 2.686\,(m)/(s)$ , c) $v_(o) \approx 7.002\,(m)/(s)$

Step-by-step explanation:

a) The maximum height reached by the rock is:

$\Delta y = (\left(0\,(m)/(s) \right)^(2)-\left(7.5\,(m)/(s) \right)^(2))/(2\cdot \left(-9.807\,(m)/(s^(2)) \right))$

$\Delta y = 2.868\,m$

$y = 1.5\,m + 2.868\,m$

$y = 4.368\,m$

Yes, the rock will exceed the top of the wall.

b) The speed when the rock reaches the top of the wall:

$v = \sqrt{\left(7.5\,(m)/(s) \right)^(2)+2\cdot \left(-9.807\,(m)/(s^(2)) \right)\cdot (4\,m-1.5\,m)}$

$v\approx 2.686\,(m)/(s)$

c) The initial speed required so that stone does not exceed the top of the wall is:

$v_(o) = \sqrt{\left(0\,(m)/(s) \right)^(2)-2\cdot \left(-9.807\,(m)/(s^(2)) \right)\cdot (4\,m-1.5\,m) }$

$v_(o) \approx 7.002\,(m)/(s)$

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Answers

i think it’s option 3, 1/3

- 5 + 7k = -19
Help pwease

Answers

Answer:

k = -2

Step-by-step explanation:

-5 + 7k = -19

7k = -14

k = -2

5. MAPPING Ben and Kate are making amap of their neighborhood on a piece of
graph paper. They decide to make one
unit on the graph paper correspond to
100 yards. First, they put their homes
on the map as shown below.

a. How many yards apart are Kate's and Ben's homes?

b. Their friend Jason lives exactly halfway between Ben and Kate. Mark the location of Jason's home on the
map. if my answers are wrong, correct me. i just used a pdf and idk if they’re right. ty ❤️

Answers

Answer:

a. 1,000 yards

b. (2, -1)

Step-by-step explanation:

Coordinates of Ben's House = (6, 2)

Coordinates of Kate's House = (-2, -4)

a. Number of yards between Kate and Ben's House using instance formula, $d = √((x_2 - x_1)^2 + (y_2 - y_1)^2)$ .

$d = √((6 -(-2))^2 + (2 -(-4))^2)$

$d = √((8)^2 + (6)^2)$

$d = √(64 + 36)$

$d = √(100)$

$d = 10 units$

1 unit = 100 yards

Therefore, 10 units = 1,000 yards.

Kate and Ben's House are 1,000 yards apart.

b. Since Jason's home is halfway between Ben and Kate's House, therefore, Jason's location is the midpoint between Ben and Kate's House. Use the midpoint formula to calculate the coordinates of Jason's location on the map, using (6, 2) and (-2, -4).

$M((x_1 + x_2)/(2), (y_1 + y_2)/(2))$

Plug in the values

$M((6 +(-2))/(2), (2 +(-4))/(2))$

$M((4)/(2), (-2)/(2))$

$M(2, -1)$

Two cars are traveling at 40 and 50 miles per hour, respectively. If the second cars out 5 miles behind the first car. How long will it take the second car to overtake the first car.

Answers

Hello There,

distance = rate x time
5 + 40x = 50x
5 = 10x
x = 1/2 hr

Hope this helps

When factoring x2 – 3xy + 2y2 by grouping the first step can be:Question 7 options:

x2 – 4xy + xy + 2y2

x2 – 2xy – xy + 2y2

x2 – 3xy + xy + 2y2

x2 + 2xy - 5xy + 2y2