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A Football Is Kicked At An Angle 37 Above The Horizontal With A Velocity Of 20.0 M/s. Calculate A

A football is kicked at an angle 37 above the horizontal with a velocity of 20.0 m/s. calculate a)the maximum?

As the ball rises to its maximum height, its vertical velocity decreases from the initial vertical velocity to 0 m/s at the rate of 9.8 m/s each second. As the ball falls from to s maximum height, its vertical velocity decreases from the initial vertical velocity to 0 m/s at the rate of 9.8 m/s each second. So, the time for the downward trip is equal to the time for the upward trip. So, the total time is equal to 2 * the time up.

vf = vi + a * t, t = (vf – vi) ÷a
vf = 0 m/s
Initial vertical velocity = 20 * sin 37 ≈ 12.0363 m/s
a = -9.8 m/s^2
Time up = (0 – 20 * sin 37) ÷ -9.8
Time up = 20 * sin 37 ÷ 9.8
The time up is approximately 1.228 seconds

d = ½ * (vi + vf) * t
d = ½ * (20 * sin 37 + 0) * (20 * sin 37 ÷ 9.8)
d = ½ * (20 * sin 37)^2 ÷ 9.8
The distance is approximately 7.39 meter. This is the maximum height.

Total time = 2 * 20 * sin 37 ÷ 9.8 = 40 * sin 37 ÷ 9.8
The total time is approximately 2.456 seconds.


Horizontal distance = Range
Range = v^2/g * sin 2θ
Range = 20^2/9.8 * sin 74
The range is approximately 39.235 meters

At the maximum height, the vertical velocity is 0 m/s. So, the velocity at the maximum height is the horizontal velocity. (20 *cos 37)

I hope this helps you understand how to solve this type of problem.

Distance up = ½ * v * sin θ * t
Time up = v * sin θ ÷ 9.8

Distance up = (½ * v * sin θ) * (v * sin θ ÷ 9.8)
Distance up = (v * sin θ)^2 ÷ 19.6
This is an easy way to determine the maximum height.

A football is kicked at a 75° angle with a velocity of 30m/s. Calculate.?

V = 30 m/s
P = 75 degrees
g = 9.8 m/s/s

a. Vx = V*cos(P)
Vy = V*sin(P)

b. 0 = Vy^2 - 2*g*H
Solve for max height H.

c. 0 = Vy*T - 1/2*g*T^2
Solve for total time T, ignoring T=0.

d. R = Vx*T
Solve for range R.

A football is kicked at an angle of 37 degrees with the horizontal with a velocity of 20.0m/sec.?

c) based on zero air resistence and uniform gravity, the velocity when it hits the ground is 20.0m/s (no friction to stop or slow it). The rest of the questions are based on vectors and trig function, but I can't remember them

A ball is projected with a velocity of 80 m/s and a horizontal angle of 30. What will the range be?

This is one of the glorious simplifications in the Physics way of thinking.It goes like this: split the velocity into two: the horizontal component and the vertical component.Focus on the vertical component: there is a well-known equation for the time of flight of particle of constant acceleration in a gravity field.THEN, given the elapsed time of flight AND the horizontal velocity work out the distance. This is the range. Or at least, it would be, if the atmosphere did not provide a resistance to movement proportional to speed squared.But as Physics deals with spherical cows, and atmospheres with vacuum like properties, let’s keep it simple.Vertical speed: 80 sin 30Horiz speed: 80 cos 30Time equation of constant acceleration:v = u + a.t where v final (vertical) velocity, u initial (vertical) velocity, a acceleration, t time. Here, u = 80 sin 30 and v = -80 sin 30 (what goes up, must come down!) and a = 9.8 m/s in SI units, so at = v - u and t = (v-u)/a. So we know t the time of flight.Then the range (on the physicist’s flat Earth which has a vacuum like atmosphere) is velocity times time, or in terms of symbols:Range = 80 cos 30 x ( v - u ) / aAnd if you recall or check with your calculator that cos 30 = 0.8660 and sin 30 = 0.5, then you have all the values needed to plug into this equation.But if you have engineering rather than physics instincts, you may ask what about the real world? Good question. The Earth is big enough that curvature is ignorable (even by an engineer) for short ranges. In 100 meters, the extra drop is 0.8 mm for example. But that air drag term. That’s a headache. You would need to know the wind speed and direction, and the drag for that shape and size at that speed - which is changing continuously. So it turns out, you would need the ball’s size too. In practice the engineer grabs his computer, and works out the speed and drag for the first fraction of a second, then uses that result to work the speed and drag in the next fraction of a second in a repeating loop. And that my dear questioner, is how we computed the trajectory of the first (and all following) Moonshots!

Football kicked at an angle 37 degrees with velocity of 20 m/s?

Break it down into horizontal and vertical speeds.
Vertical is 20sin37 = 12.04 m/s
Horizontal is 20cos37 = 15.97 m/s

take the vertical first.
h = v²/2g = 7.396 meters
t = √(2d/g) = 1.229 seconds.
total time to landing is twice that or 2.46 seconds

Horizontally, how far does it travel in 2.46 seconds
d = 15.97 m/s x 2.46 s = 39.2 feet

Max height is 7.396 meters, and at that point, acceleration is downward at 9.8 m/s², as it is through out the flight, and velocity is horizontal at 15.97 m/s

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A boy kicks a football with an initial velocity of 28.0 m/s at an angle of 30.0° above the horizontal. What?

A boy kicks a football with an initial velocity of 28.0 m/s at an angle of 30.0° above the horizontal. What is the highest elevation reached by the football in its trajectory?
A) 11.2 m
B) 10.0 m
C) 12.7 m
D) 9.40 m
E) 41 m

9. A boy kicks a football with an initial velocity of 20 m/s at an angle of 25 above the horizontal...?

A boy kicks a football with an initial velocity of 20 m/s at an angle of 25 above the horizontal. The acceleration of the ball while it is in flight is?

A. 20 m/s2.
B. -20 m/s2.
C. -9.8 m/s2.
D. 0 m/s2.

A football is kicked at 37 degrees to the horizontal at 20.0m/s from the player's hand at 1.00m from the groun?

Let’s determine the vertical and horizontal components of the initial velocity.
Vertical = 20 * sin 37, Horizontal = 20 * cos 37

The vertical velocity decreases at the rate of 9.8 m/s each second. The horizontal velocity is constant. Use the following equation to determine the time the ball is in the air.

d = vi * t + ½ * a * t^2
d is the ball’s vertical displacement.
Vertical displacement = Final height – Initial height = 0 – 1 = -1 m
vi = 20 * sin 37, a = -9.8

-1 = 20 * sin 37 * t + ½ * -9.8 * t^2
4.9 * t^2 – 20 * sin 37 * t – 1 = 0

I use the following website to solve quadratic equations.
http://www.math.com/students/calculators...
t = 2.53683518503705
Let’s round to 2.537 seconds

Use the following equation to determine the horizontal distance the ball travels.
d = v * cos θ * t
d = 20 * cos 37 * 2.537
The distance is approximately 40.52 meters.

A football was kicked at 22.5 m/s at an angle of 35 degrees above the horizontal. What was the ball's hangtime?

22.5m/s at 35° means the vertical ball speed (ascending) is sin(35°) x 22.5m/s = 12.9m/s.The gravity force is now accelerating the ball downward at 9.81m/s². We can deduce the time the ball will stop going up and start going down by doing 12.9/9.81 = 1.315 s.Now we know it takes about 1.315 second for the ball to stop going up. As the acceleration rate is constant, the ball will take the same time to reach the ground.So the total flight time of the ball will be about 2 x 1.315 = 2.63 seconds.Of course this result do not include height difference between the starting point and the landing point, neither the energy loss due to friction with air. To take friction into account, we will need the mass and the diameter of the ball.

A ball is projected with a velocity of 15m/s making an angle of 30 degree with the horizontal. What is the time of flight of the ball (given g=10m/s)?

Q: A ball is projected with a velocity of 15m/s making an angle of 30 degree with the horizontal. What is the time of flight of the ball (give g=10m/s)?A: If we take the usual easy-out of no air drag, then using the splendid physics student’s method of working the horizontal and vertical axes separately, we soon realize that the total time of flight is controlled by the vertical axis: so when the ball descends to the starting height, the game is over!The only other trick needed is elementary trigonometry. The vertical speed is the vertical side of a triangle, whose diagonal represents the velocity of 15 m/s, using the formula for the sine ratio, which is vertical divided by diagonal. So we calculate the length of the vertical (representing the vertical speed) in this way:sine 30 deg = (vertical / diagonal)= (vertical speed / 15)so vertical speed = 15*sin(30)= 15*0.5 = 7.5 m/sNow we chose the applicable equation from the four main kinematic equations:We know the initial vertical velocity, we can easily see that the constant acceleration (g) reduces the velocity equally over time and so the ball lands with the same speed it started - in the other direction!V final = V initial + a*t will work for us if we visualize the trajectory carefully; using up is positive and g is negative we get -7.5 = 7.5 -10*t which simplifies to 10*t = 15 and t = 1.5 seconds.

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