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When a rock thrown straight up climbs for 3.00 s before falling. Neglecting air resistance with what velocity did the rock strike the ground?
You can use Newton's equations of motion: At the top of the climb its velocity u = 0 m/s Its acceleration is acceleration due to gravity a ≈ 9.8 m/s Time of descent t = time of ascent = 3.00 s (I'll assume positive is towards the ground) v = u + at ≈ 0 m/s + 9.8 m/s² × 3.00 s = 29.4 m/s HOWEVER, this is the velocity (towards the ground) reached when the rock has returned to height from which it was thrown (released) above the ground - unless the rock was "thrown" by an explosive force at ground level, the rock will not have reached the ground at this point: there is still the distance from which it was "thrown". Which means its final velocity at ground level can be found using: v² = u² + 2as v = velocity it hits the ground u ≈ 29.4 m/s (as found above) s = distance above ground from which the rock was "thrown" = height_of_throw m a = acceleration due to gravity ≈ 9.8 m/s → v² = u² + 2as → v ≈ √((29.4 m/s)² + 19.6 m/s² × height_of_throw m) = √(864.36 + 19.6 × height_of_throw) m/s
A rock is thrown upward at an angle of 50 degree with respect to the horizontal As it risesm its horizontal component of velocity?
If you ignore the effect of the air grabbing at it and only figure in gravity, then the horizontal component of velocity is constant, from the time the stone leaves your hand until the time it hits the ground. Makes no difference whether you toss it up, down, horizontal, or on a slant. Also makes no difference whether it's a cannonball, a stone, or a bullet.
A ball is thrown straight up At the top of its path its instantaneous speed is?
If it is thrown at an angle, at the top of its path, its vertical velocity will be zero, however its horizontal velocity will be the same as its initial horizontal velocity minus whatever loss in speed as a result of air friction at that point. We won't know what that is without more information.
Does vertical velocity effect range?
We're guessing that it does.We've noticed that a rock thrown at some angle abovehorizontal will go farther thanan identical rock thrown at an identical speed and at the same angle below horizontal.Major league outfielders seem to know this, either by virtue of some primitive instinctor else from practice and experience. When one of them attempts a long throw to Home,they never throw it down, always up.
What is the vertical velocity at the top of the path of a projecticle thrown striaght up or launched at an angle?
The vertical velocity at the top of the path of a projectile thrown straight up is 0 m/s because it momentarily stops before falling back down. For a projectile launched at an angle, the vertical velocity at the top of the path depends on the initial velocity and launch angle, but it will also momentarily be 0 m/s before changing direction.
When a rock thrown straight up climbs for 3.00 s before falling. Neglecting air resistance with what velocity did the rock strike the ground?
You can use Newton's equations of motion: At the top of the climb its velocity u = 0 m/s Its acceleration is acceleration due to gravity a ≈ 9.8 m/s Time of descent t = time of ascent = 3.00 s (I'll assume positive is towards the ground) v = u + at ≈ 0 m/s + 9.8 m/s² × 3.00 s = 29.4 m/s HOWEVER, this is the velocity (towards the ground) reached when the rock has returned to height from which it was thrown (released) above the ground - unless the rock was "thrown" by an explosive force at ground level, the rock will not have reached the ground at this point: there is still the distance from which it was "thrown". Which means its final velocity at ground level can be found using: v² = u² + 2as v = velocity it hits the ground u ≈ 29.4 m/s (as found above) s = distance above ground from which the rock was "thrown" = height_of_throw m a = acceleration due to gravity ≈ 9.8 m/s → v² = u² + 2as → v ≈ √((29.4 m/s)² + 19.6 m/s² × height_of_throw m) = √(864.36 + 19.6 × height_of_throw) m/s
A rock is thrown upward at an angle of 50 degree with respect to the horizontal As it risesm its horizontal component of velocity?
If you ignore the effect of the air grabbing at it and only figure in gravity, then the horizontal component of velocity is constant, from the time the stone leaves your hand until the time it hits the ground. Makes no difference whether you toss it up, down, horizontal, or on a slant. Also makes no difference whether it's a cannonball, a stone, or a bullet.
What is a cone shaped mountain formed out of rock thrown up from inside the earth called?
A cone-shaped mountain formed from rock thrown up from inside the Earth is called a volcano.
If a ball is thrown vertically up with the velocity of 40m s what would be its velocity after two seconds?
Ignoring air resistance and using g = 9.81 ms-2, velocity = 20.38 ms-1.
A balls is thrown upwards with a velocity of 10 ms How high does it go?
The ball goes up 5.1 meters.
A ball is thrown straight up At the top of its path its instantaneous speed is?
If it is thrown at an angle, at the top of its path, its vertical velocity will be zero, however its horizontal velocity will be the same as its initial horizontal velocity minus whatever loss in speed as a result of air friction at that point. We won't know what that is without more information.
Does vertical velocity effect range?
We're guessing that it does.We've noticed that a rock thrown at some angle abovehorizontal will go farther thanan identical rock thrown at an identical speed and at the same angle below horizontal.Major league outfielders seem to know this, either by virtue of some primitive instinctor else from practice and experience. When one of them attempts a long throw to Home,they never throw it down, always up.
A body can escape the gravitational pull of the earth if it is thrown up with a velocity of?
In order for a body to escape the gravitational pull of the Earth, it needs to be thrown up with an initial velocity equal to or greater than the escape velocity of around 11.2 km/s. This velocity allows the object to overcome the gravitational pull of the Earth and continue traveling away from it indefinitely.
Can an object ever have a velocity of zero if it moves from starting location?
Yes. An example of this would be a ball thrown straight up; at the very peak of its travel, it's not moving either up or down and therefore has an instantaneous velocity of zero.
When a rock is thrown up into the air it reaches it highest point and briefly comes to a complete stop before it starts to fall back to the ground which is greatest at the point where the rock stops?
At the highest point, the rock's velocity is momentarily zero as it changes direction from going up to coming back down. At this point, the rock has the maximum potential energy, which is then converted into kinetic energy as it falls back towards the ground. The greatest potential energy occurs at the highest point of the rock's trajectory.
What is the boundary above which objects thrown up do not reach the Earth called?
The limit is not so much a distance from Earth, but rather a velocity - called the escape velocity. (roughly 25000 mph) /Brian W
What is the vertical velocity at the top of the path of a projecticle thrown striaght up or launched at an angle?
The vertical velocity at the top of the path of a projectile thrown straight up is 0 m/s because it momentarily stops before falling back down. For a projectile launched at an angle, the vertical velocity at the top of the path depends on the initial velocity and launch angle, but it will also momentarily be 0 m/s before changing direction.
