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11Alexandr11 [23.1K]
3 years ago
12

A soccer ball is kicked horizontally and rolled off of a 30 meter high cliff. The soccer ball was kicked with a velocity of 17m/

s. How long did it take before the ball hit the ground? How far from the cliff did the ball land? (Make sure to include units.)
VERTICLE
Acceleration =
Distance =
Initial velocity=
Time=
HORIZONTAL
Distance=
Velocity=
Time=
Show your work when solving for time:


Show your work when solving for horizontal distance:
Physics
1 answer:
igor_vitrenko [27]3 years ago
4 0

Answer:This will hopefully help. It took me a while so will you mark me brainliest? Thanks!

Explanation:

velocity=Vertical (y) info:

• y = -22.0 m

• viy = 0 m/s

• ay = -9.8 m/s2

distance=

acceleration=

time=

horizontal=Horizontal (x) info:

• x = 35 m

• vix = ?

• ax = 0 m/s2

Initial velocity=16.5 m/s

VOCAB:Horizontal motion is constant

velocity- motion

acceleration-increase in the rate or speed of something.

distance-Distance is a numerical measurement of how far apart objects or points are. In physics or everyday usage, distance may refer to a physical length or an estimation based on other criteria (e.g. "two counties over"). The distance from a point A to a point B is sometimes denoted as. .

time-a point of time as measured in hours and minutes past midnight or noon.

Explanation:

The soccer ball travels a horizontal distance of

33.9 m after being kicked with horizontal velocity of 18.8ms−1. Assuming air resistance is negligible.

Time of flight of the ball

t=Distance Speed=33.918.8s In the vertical direction kinematic equation is

height

h=ut+12gt2 Initial velocity uv=0, acceleration due to gravity g=9.81ms−2. Inserting various values we get h=0×33.918.8+12×9.81×(33.918.8)2⇒h=12×9.81×(33.918.8)2

⇒h=15.9m, rounded to 3 sig. figs

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A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
3 years ago
2. Use the diagram below to answer this question. As the ball moves from point A
Thepotemich [5.8K]

Answer:

at point A the ball possess pontetial energy , point B kinetic energy then point C pontetial energy

5 0
3 years ago
Please help ill mark youas brainliest !!
USPshnik [31]

Answer:

can you put on a clearer image this one is hard to see

8 0
2 years ago
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

7 0
3 years ago
You make a straight line when an object is
swat32

Answer:

can you explain more or..,.

Explanation:

8 0
4 years ago
Read 2 more answers
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