Effect of speed on horizontal displacement in projectile motion

AI Thread Summary
In projectile motion, the horizontal displacement is not directly proportional to the speed; instead, it is proportional to the square of the launch speed when the launch angle remains constant. Halving the speed results in a range that decreases by a factor of one-fourth, not half, due to the relationship between speed, angle, and energy. The time of flight remains constant if the launch height is unchanged, but the horizontal component of velocity must be considered separately. The equation Δx = vΔt applies to one-dimensional motion, and in two-dimensional motion, the horizontal velocity component must be analyzed. Understanding the distinction between overall speed and horizontal speed is crucial for accurate calculations in projectile motion.
darla1608
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Homework Statement



Question: Provided all else is equal, if a ball is thrown at half the original speed will it travel half as far?


Homework Equations



Δx = vΔt


The Attempt at a Solution



In projectile motion, the time of flight will be the same no matter the horizontal velocity, as long as the height is the same. When I plug two different velocities into the above equation ( one being half of the other), the displacement is half.

Issue with this? The answer in the book states this is not true. What am I doing wrong or not considering?
 
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darla1608 said:

Homework Statement



Question: Provided all else is equal, if a ball is thrown at half the original speed will it travel half as far?

Homework Equations



Δx = vΔt

The Attempt at a Solution



In projectile motion, the time of flight will be the same no matter the horizontal velocity, as long as the height is the same. When I plug two different velocities into the above equation ( one being half of the other), the displacement is half.

Issue with this? The answer in the book states this is not true. What am I doing wrong or not considering?

Range is proportional to the square of the launch speed and to the sine of twice the launch angle.

If you keep the same angle but halve the speed, the range will be decrease by a factor of 1/4.

What you appear to be trying to do is launch the projectile with the same vertical speed (so it will reach the same height) but less horizontal speed. This means it will have lower launch energy and a larger launch angle (relative to the horizontal).

AM
 
Last edited:
Let Δt be the time it takes for the projectile to travel.

The equation Δx = vΔt only applies for one-dimensional kinematics. In 2D kinematics, the v is the x component of velocity. Just because velocity has been halved doesn't mean that the x component of velocity has been halved.
 
thank you
 
"Speed" is not the same as "horizontal component of speed". To go from one to the other you need to know the angle the trajectory makes with the horizontal.
 

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