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Gre4nikov [31]
2 years ago
7

The height of a projectile launched upward at a speed of 25 meters/second from a height of 70 meters is given by the function h(

t)=-5sup(t,2)+25t+70. How long will it take the projectile to hit the ground?
Mathematics
1 answer:
N76 [4]2 years ago
5 0

6.5 s  take the projectile to hit the ground.

<h3>What is projectile?</h3>

A projectile is any object thrown into space upon which the only acting force is gravity.

Given:

speed = 25 m/s

h(t) = -5t² + 25t  + 70

height = 70 m

Time to reach maximum height

( u sin90)/2g

= 25/2*9.8

= 1.275s

Max height = u²Sin²\theta/g

=25*25/9.8

=63.775

So, total height=63.775 + 70 = 133.775 m

Now, time to fall back to the ground

h = ut+(1/2)gt²

t² = (h- ut)2/g

u=0

H= 133.775 m

(133.775*2)/9.8= t²

t² = 27.3010

t= 5.225 s

Total time taken to hit ground

= 5.225 + 1.275

= 6.5 s

Learn more about this concept here:

brainly.com/question/17096336

#SPJ1

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25, 2, and 3 are rational.
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Helga [31]

Answer:

Since the Line is Perpendicular

m.m'=-1

The line

y=x+8

Comparing with

Y=mx + C

m=1

m.m'=-1

m'=-1/m = -1/1 = -1.

y-y' = m'(x-x')

The point it passes is (-7,8)

y-8= -1(x--7)

y-8=-1(x+7)

y-8 = -x - 7

y + x -1 = 0.

or In Slope intercept Form

Just Make y the subject

y= -x + 1 ..... This is your answer.

Hope this helps!!!

6 0
2 years ago
Please solve this question!!​
GaryK [48]

Answer:  (i) 1/221     (ii) 11/221      (iii) 95/663        (iv) 1/663

<u>Step-by-step explanation:</u>

(i) A deck of cards contains 4 Kings out of 52 total cards

1st draw: 4 Kings out of 52 total cards → 4/52 = 1/13

2nd draw: 3 remaining Kings out of 51 total remaining cards  →  3/51 = 1/17

  <u>1st Draw </u>                   <u>2nd Draw </u>                  <u>Outcome</u>         <u>Probability</u>

 King: P(K) = 1/13        King: P(K₂/K₁) = 1/17     King, King       (1/13) x (1/17) = 1/221

*************************************************************************************************

(ii) A deck of cards contains 4 Jacks, 4 Queens, & 4 Kings out of 52 total cards

1st draw: 12 Face cards out of 52 total cards → 12/52 = 3/13

2nd draw: 11 remaining Face cards out of 51 total remaining cards  →  11/51

<u>1st Draw </u>                 <u>2nd Draw </u>                  <u>Outcome</u>       <u>Probability</u>

 Face: P(F) = 3/13    Face: P(F₂/F₁) = 11/51   Face,Face     (3/13) x (11/51) = 11/221

*************************************************************************************************

(iiI) A deck of cards contains 26 black cards out of 52 total cards but there are 2 black Jacks, 2 black Queens, and 2 black Kings.

1st draw: 20 Black (not Face) cards out of 52 total cards → 20/52 = 5/13

2nd draw: 19 remaining Black (not Face) cards out of 51 total remaining cards  →  19/51

  <u>1st Draw </u>                 <u>2nd Draw </u>                       <u>Outcome</u>   <u>Probability</u>

Black: P(B~) = 5/13   Black: P(B~₂/B~₁) = 19/51    B~,B~    (5/13) x (19/51) = 95/663

*************************************************************************************************

(ii) A deck of cards contains 4 Aces out of 52 total cards

1st draw: 4 Aces out of 52 total cards → 4/52 = 1/13

2nd draw: 1 Queen of Hearts out of 51 total remaining cards  →  1/51

  <u>1st Draw </u>                 <u>2nd Draw </u>                   <u>Outcome</u>         <u>Probability</u>

 Ace: P(A) = 1/13      Qh: P(Qh₂/A₁) = 1/51   Ace,Queen(h)   (1/13) x (1/51) = 1/663

7 0
4 years ago
The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m st
Reil [10]

Answer:

a) 30.726m/s and b) 5.5549s

Step-by-step explanation:

a.) What was Chris Huber’s speed in meters per second(m/s)?

Given the distance and time, the formula to obtain the speed is

v=\frac{d}{t}.

Applying this to our problem we have that

v=\frac{200m}{6.509s}= 30.726m/s.

So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.

b) What was Whittingham’s time through the 200 m?

In a) we stated that v=\frac{d}{t}. This formula implies that

  1. t=\frac{d}{v}.

First, observer that 19\frac{km}{h} =19,000\frac{m}{h}=\frac{19,000}{3,600}m/s= 5.2777m/s.

Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, v=30.726\frac{m}{s} +5.2777\frac{m}{s}= 36.003 m/s.

Then, applying 1) we have that

t=\frac{200m}{36.003m/s}=5.5549s.

So, Sam Whittingham’s time through the 200 m was 5.5549s.

5 0
3 years ago
Zero is an element of the set of natural numbers
kipiarov [429]
That is true. it is part of natural numbers.
4 0
3 years ago
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