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Q: Jules kicks a soccer ball off the ground and in the air with an initial velocity of 25 feet per second Approximately what maximum height does the soccer ball reach?

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initial velocity of the kick = 28.06 m/s

The object's initial distance above the ground The object's initial velocity

The answer will depend on what "it" is, and on what its initial velocity is.

No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.

To answer this question one would need to know the rock's initial height and velocity.

Perhaps you mean Terminal Velocity, as in a parachute fall? This is the maximum speed reached in the fall. Final velocity will be zero, assuming you arrive on the ground.

20.40

initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)

This is a velocity question so u need to use uvaxt

when the object reaches maximum height, the velocity of the object is 0 m/s.It reaches maximum height when the gravity of the body has slowed its velocity to 0 m/s. If there is no gravity and there is no external force acting on it then it will never reach a maximum height as there wont be a negativeaccelerationdemonstrated by newtons first law.Where there is and you have the objects initial velocity then you can use :v^2 = u^2+2.a.sv = Velocity when it reaches Max. height so v = 0u = Initial Velocity (m/s)a = Retardation/ Negative Acceleration due to gravity, -9.80m/s ^2And then the unknown (s) is the displacement, or height above ground, and if everything else is in the right format it should be in metres.

Perhaps you mean terminal velocity. This is the maximum velocity reached by an object falling to the ground when the acceleration due to gravity is matched by the drag resistance of the air through which it is falling.

The answer depends on its initial velocity and the height from which its fall to the ground is measured.

51 seconds.

51 seconds

For a given velocity, the maximum distance will be achieved when the projectile is launched at an angle of 45o (neglecting air resistance).

i will give u an illustration, consider an object projected (thrown)with some initial vertical velocity from the ground such that it traces a open downward parabolicpath, in that path the vertical displacement of the body from the point of projection to the point where it strikes the ground is equal to zero,but it have some velocity.

If a shell is fired from the ground with velocity of 1600 m and an angle of 64 to the horizontal then it would have a horizontal rang of 55.0. This is considered math.

10m/s

if a body is thrown having initial velocity and make angle with ground this body is known as projectile and the way is calle trajectory

algebra 2 right? i hated that unit man i forgot everything we learned in that class

If you're willing to ignore the effect of air resistance, then the answer is as follows: The object's horizontal velocity remains constant (at least until it eventually hits the ground). The vertical component of the object's initial velocity ... call it V(i) ... is the (total initial velocity) multipled by the (sine of the initial angle above the horizontal). Beginning at the time of the toss, the magnitude of the vertical component of velocity is V = V(i) - 1/2gT2. T = number of seconds after the toss g = acceleration of gravity = approx 32 ft/sec2 or 9.8 m/sec2

Using the acceleration formula, final acceleration is the final velocity minus the initial velocity over elapsed time. Final velocity you gave as 40m/s, and the initial velocity was zero (the apple was stationary on the tree), so the difference is 40 m/s. Divided by the time you gave, 4 s, this will be 10 m/s²

This result is because the wet ball carries more inertia to weight ratio before hitting the ground , it then compresses, loses some of the liquid weight, becomes lighter, and because of the initial inertial force, can therefore leave the ground at a greater velocity

assuming uniform decelaration, . number of g's = ((v^2 - u^2) / (2 * s)) / 9.82 v = final velocity u = initial velocity s = distance

The initial velocity of a dropped ball is zero in the y (up-down) direction. After it is dropped gravity causes an acceleration, which causes the velocity to increase. F = ma, The acceleration due to gravity creates a force on the mass of the ball.

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