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nydimaria [60]
3 years ago
8

It is the final seconds of an ice hockey game between the Flyers and the Bruins. The Bruins are down by 1 point. With 20 s left

in the game, the Bruins pull the goalie and have him play as a forward in an attempt to tie the game.
The Flyers successfully defend their goal for 9 s. With only 1.25 s remaining on the game clock, a Flyer shoots the puck on the ice past the skates and sticks of the other players and toward the Bruins' goal. The puck is 37 m from the goal when it leaves the stick with an initial horizontal velocity of 30 m/s. The shot is perfectly directed toward the empty goal, but the ice slows the puck down at a constant rate of 0.50 m/s2 as it slides toward the goal. None of the Bruins can stop the puck before it reaches the goal.
a. Where is the puck when the game clock reaches zero and the horn sounds to end the game?
b. Do the Flyers win the game by 1 or 2 points?
Physics
1 answer:
liberstina [14]3 years ago
6 0

Answer:

a

P =0.1094 \ m past the goal post

Yes the Flyers won the match

Explanation:

Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as

     D =  ut + \frac{1}{2} * a * t^2

=>  D =30  * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2

=>   D =37.1094 \ m

Generally the position of the puck when the game clock reaches zero is mathematically represented as

     P = 37.1094 -37

       P =0.1094 \ m past the goal post

Given that the Bruins where one point down and Flyers scored another goal  it means that the flyer are now two point up hence they won the match

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Answer:

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A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav
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Answer:

a) The takeoff speed is 10 m/s.

b) The maximum height above the ground is 1.2 m.

Explanation:

The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v =(v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

α = jumping angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time "t"

a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:

r final = (8.7 m, 0 m)

Then at final time:

8.7 m = x0 + v0 · t · cos α

0 m = y0 + v0 · t · sin α + 1/2 · g · t²

(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:

8.7 m = v0 · t · cos α

Solving for "v0":

8.7 m /(t · cos α) = v0

Replacing v0 in the equation of the y-component, we can obtain the final time:

0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)

- 8.7 m · tan 29° / -4.9 m/s² = t²

t = 0.99 s

Now, we can calculate the initial speed:

8.7 m /t · cos α = v0

v0 = 8.7 m / (0.99 s · cos 29°)

<u>v0 = 10 m/s</u>

The takeoff speed is 10 m/s

b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:

0 = v0 · sin α + g · t

Solving for "t":

-v0 · sin α / g = t

t = - 10 m/s · sin 29° / 9.8 m/s²

t = 0.49 s

Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.

Now, let´s find the height of the kangaroo at that time:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²

<u>y = 1.2 m</u>

The maximum height above the ground is 1.2 m.

4 0
3 years ago
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inysia [295]
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Answer:

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First we will find the speed of the ball just before it will hit the floor

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