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Chapter 3: Problem 7

Is a projectile's speed constant throughout its parabolic trajectory?

Short Answer

Expert verified
No, the speed of a projectile is not constant throughout its parabolic trajectory. It changes due to the continuous acceleration caused by gravity.

Step by step solution

01

Understanding Projectile Motion

Recognize that when a projectile is launched, it follows a parabolic trajectory. The motion of the projectile can be divided into horizontal and vertical components. The horizontal component of velocity remains constant throughout if air resistance is neglected. This is because there are no horizontal forces acting on the projectile after launch. On the other hand, the vertical component of the velocity continuously changes due to the force of gravity constantly pulling the projectile downwards.
02

Role of Gravity

Understand the role of gravity in projectile motion. Gravity, a downward force, constantly accelerates the projectile towards Earth. This means that the vertical component of the projectile's velocity is continuously changing - it decreases as the projectile ascends, becomes zero at the maximum height, and then increases as the projectile descends.
03

Analyzing the Speed

Analyze the speed of the projectile. The speed of the projectile is the magnitude of its velocity vector. Since the vertical velocity component of the projectile is changing due to gravity (while the horizontal component is constant), the speed of the projectile (the magnitude of the velocity vector) is not constant throughout the trajectory. It is at its maximum at launch and impact (when the trajectory angle is not zero), and it is at its minimum (but not zero) at the maximum height of the projectile.

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Most popular questions from this chapter

An object undergoes acceleration \(2.3 \hat{\imath}+3.6 \hat{\jmath} \mathrm{m} / \mathrm{s}^{2}\) for \(10 \mathrm{s}\). At the end of this time, its velocity is \(33 \hat{\imath}+15 \hat{\jmath} \mathrm{m} / \mathrm{s} .\) (a) What was its velocity at the beginning of the 10 -s interval? (b) By how much did its speed change? (c) By how much did its direction change? (d) Show that the speed change is not given by the magnitude of the acceleration multiplied by the time. Why not?

You wish to row straight across a 63 -m-wide river. You can row at a steady \(1.3 \mathrm{m} / \mathrm{s}\) relative to the water, and the river flows at \(0.57 \mathrm{m} / \mathrm{s} .\) (a) What direction should you head? (b) How long will it take you to cross the river?

A kid fires a squirt gun horizontally from \(1.6 \mathrm{m}\) above the ground. It hits another kid \(2.1 \mathrm{m}\) away square in the back, \(0.93 \mathrm{m}\) above the ground. What was the water's initial speed?

Is there any point on a projectile's trajectory where velocity and acceleration are perpendicular?

The table below lists position versus time for an object moving in the \(x-y\) plane, which is horizontal in this case. Make a plot of position \(y\) versus \(x\) to determine the nature of the object's path. Then determine the magnitudes of the object's velocity and acceleration. $$\begin{array}{ccc} \text { Time, } t(\mathrm{s}) & x(\mathrm{m}) & y(\mathrm{m}) \\ 0 & 0 & 0 \\ 0.10 & 0.65 & 0.09 \\ 0.20 & 1.25 & 0.33 \\ 0.30 & 1.77 & 0.73 \\ 0.40 & 2.17 & 1.25 \\ 0.50 & 2.41 & 1.85 \\ 0.60 & 2.50 & 2.50 \end{array}$$ $$\begin{array}{ccc} \text { Time, } t(\mathrm{s}) & x(\mathrm{m}) & y(\mathrm{m}) \\ 0.70 & 2.41 & 3.15 \\ 0.80 & 2.17 & 3.75 \\ 0.90 & 1.77 & 4.27 \\ 1.00 & 1.25 & 4.67 \\ 1.10 & 0.65 & 4.91 \\ 1.20 & 0.00 & 5.00 \end{array}$$
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