<span>the deeper you go the more pressure...that's for depth</span>
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:
spring deflection is x = (v2 / R + g) m / 4
Explanation:
We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration
Let's write the equations on the Y axis of this description
Fe - W = m 
Where Fe is elastic force, W the weight and
the centripetal acceleration. The elastic force equation is
Fe = - k x
4 (k x) - mg = m v² / R
The four is because there are four springs, R is theradio of dip
We can calculate the deflection (x) of the springs
x = (m v2 / R + mg) / 4
x = (v2 / R + g) m / 4
Answer:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
Sound waves....................................