For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.
<h3>Explanation</h3>
How long does it take for the ball to reach the goal?
Let the distance between the kicker and the goal be 
meters.
Horizontal velocity of the ball will always be 
until it lands if there's no air resistance.
The ball will arrive at the goal in 
seconds after it leaves the kicker.
What will be the height of the ball when it reaches the goal?
Consider the equation
.
For this soccer ball:
,
,
since the player kicks the ball "from ground level."
when the ball reaches the goal.
.
Solve this quadratic equation for , 
.
meters when 
meters.
or 
meters when 
meters.
In other words,
- For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
- For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.
Answer:
It is frequently stated that the value of the acceleration due to gravity at the pole is larger than at the equator because the poles are closer to the center of the earth due to the earth's oblateness. ... The measured value is larger because the earth's density is not uniform but increases toward the center.
Answer:
Mechanical advantage = 15
Explanation:
Given the following data;
Output force = 3000N
Input force = 200N
To find the mechanical advantage;
Mechanical advantage = output force/input force
Substituting into the equation, we have
Mechanical advantage = 3000/200
Mechanical advantage = 15
Answer:
See below
Explanation:
<u>I will use 3 x 10^8 m/s for speed or wave</u>
speed = wavelength * frequency
3 x 10^8 = w * 7.34 x 10^2 <====== are you sure this isn't KILO Hz ?
w = <u>408719. 3 meters </u>
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
