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When A ball is thrown straight up from the ground. What way does its acceleration point at the top?
Wiki User
∙ 6y ago
Its acceleration points straight down at all times after it's released.
Wiki User
∙ 6y ago
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What is the acceleration at highest pointn when ball is thrown straight up?
The acceleration is the acceleration of gravity, downwards, or 9.8m/s/s (32 ft/s/s). When ball is thrown straight up it has an initial velocity that is decreasing because of gravity; at the highest point velocity is zero but acceleration is always constant at gravity rate.
When a ball is thrown up what is its acceleration and velocity at the highest point?
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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 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.
A stone is thrown straight up. when it reaches its highest point?
When the stone reaches its highest point, earth's gravity ensures it has to come down.
What is the acceleration at highest pointn when ball is thrown straight up?
The acceleration is the acceleration of gravity, downwards, or 9.8m/s/s (32 ft/s/s). When ball is thrown straight up it has an initial velocity that is decreasing because of gravity; at the highest point velocity is zero but acceleration is always constant at gravity rate.
An object thrown upward has zero acceleration at the highest point?
No, the acceleration at the highest point is never 0.
When a ball is thrown up what is its acceleration and velocity at the highest point?
0 zero
In an acceleration graph showing speed versus time a straight line shows the acceleration is?
A graph that shows speed versus time is not an acceleration graph.The slope of the graph at any point is the acceleration at that time.A straight line shows that the acceleration is constant.
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 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.
A stone is thrown straight up. when it reaches its highest point?
When the stone reaches its highest point, earth's gravity ensures it has to come down.
Consider a ball thrown upward It goes to its highest point and then falls until you catch it at the bottom of its flight What is the acceleration of the ball at its highest point?
0 ms-2 upwards
A ball is thrown straight up into the air at what position is its kinetic energy minimum?
At the highest point, the kinetic energy is least.
What hasa zero velocity yet a non-zero acceleration?
For example, an object thrown upwards, when it is at its highest point. This situation is only possible for an instant - if the acceleration is non-zero, the velocity changes, and can therefore not remain at zero.
What is the equation to find acceleration of an object moving in a straight line?
It is the second derivative of its distance from a fixed point on the line, with respect to time. There is nothing in the question which entitles you to assume that the acceleration is uniform.
What is the acceleration of an object effected by gravity thrown at a 45 degree angle above the horiziontal at its highest point?
The only force acting on a projectile once launched is gravity. So the acceleration of any object launched at any angle is the acceleration due to gravity, -9.8m/s2.
