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Chapter 3: Problem 70

If the time of flight of the ball is \(t\) seconds, at what point in time will the ball have zero vertical velocity? A. \(t / 4 \mathrm{s}\) B. \(t / 4 \mathrm{s}\) C. \(t / 2 \mathrm{s}\) D. \(t / 2 \mathrm{s}\) E. There is no place on the path where the vertical velocity is zero.

Short Answer

Expert verified
C. \(t / 2 \) seconds.

Step by step solution

01

Understand the Problem

We have a ball in projectile motion, and we need to determine when during its flight the ball's vertical velocity will be zero. The total time of flight is denoted by \( t \).
02

Recall Properties of Projectile Motion

In projectile motion, the vertical velocity of the ball is zero at the peak of its flight. This occurs halfway through the total time of flight because the path is symmetrical.
03

Calculate the Time to Peak Height

Since the peak occurs halfway through the motion, we simply divide the total time of flight by 2. Thus, the time when the vertical velocity is zero is given by \( \frac{t}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Velocity
Vertical velocity in projectile motion is the component of the object's velocity that is directed along the vertical axis. When a ball is thrown into the air, it initially rises, slows down, stops momentarily at the peak of its trajectory, and then descends. This changing speed is due to the influence of gravity.

  • Initial Vertical Velocity: When the ball is thrown upwards, it has a certain initial vertical velocity (\( v_{iy} \)).
  • Deceleration: As the ball moves upwards, gravity decelerates it until it reaches the highest point where the vertical velocity is zero.
  • Acceleration: After reaching the peak, gravity accelerates the ball downwards until it hits the ground or another surface.
The key takeaway is that the vertical velocity is only zero briefly at the peak of the projectile's path. At any other point in time during its flight, the ball has a non-zero vertical velocity as it moves up or down.
Time of Flight
Time of flight in projectile motion refers to the total duration for which the ball remains in the air. Understanding this concept is crucial as it helps determine other characteristics of the projectile's motion, such as when the vertical velocity reaches zero.

A few key points about the time of flight:
  • The time of flight (\( t \)) is determined by both the initial velocity of the projectile and the angle of launch.
  • In symmetrical projectile motion without air resistance, the time of ascent equals the time of descent.
To find specific moments during the flight, such as when the vertical velocity is zero, it is important to note that this happens exactly halfway through the time of flight. Hence, the time of flight provides a framework for understanding the entire motion.
Peak Height
Peak height represents the maximum vertical position a projectile reaches during its flight. At this point, the ball temporarily stops moving upwards before it starts to descend. This is directly related to vertical velocity and time of flight.

At peak height, the projectile's vertical velocity is zero, as the upward motion ceases and transitions to downward motion. This occurs because gravity has counteracted the initial upward force.

  • The time to reach peak height in symmetrical projectile motion can be calculated using the formula: \( \text{Time to peak height} = \frac{t}{2} \), where \( t \) is the total time of flight.
  • The peak height itself depends on the initial vertical velocity and can be influenced by factors such as launch angle and speed.
Understanding peak height is essential when analyzing projectile motion. It tells us not just about the maximum altitude reached but also provides insights into the symmetrical nature of projectile paths.

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Most popular questions from this chapter

Fighting forest fires. When fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. A pilot is practicing by dropping a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path 90.0 \(\mathrm{m}\) above the ground and with a speed of \(64.0 \mathrm{m} / \mathrm{s}(143 \mathrm{mi} / \mathrm{h}),\) at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

\cdotse: Leaping the river, II. A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle. (See Figure \(3.45 .\) ) The takeoff ramp was inclined at \(53.0^{\circ},\) the river was 40.0 \(\mathrm{m}\) wide, and the far bank was 15.0 \(\mathrm{m}\) lower than the top of the ramp. The river itself was 100 \(\mathrm{m}\) below the ramp. You can ignore air resistance. (a) What should his speed have been at the top of the ramp for him to have just made it to the edge of the far bank? (b) If his speed was only half the value found in \((a)\) where did he land?

. A man stands on the roof of a 15.0 -m-tall building and throws a rock with a velocity of magnitude 30.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. You can ignore air resistance. Calculate (a) the maximum height above the roof reached by the rock, (b) the magnitude of the velocity of the rock just before it strikes the ground, and (c) the horizontal distance from the base of the building to the point where the rock strikes the ground.

\bulletA test rocket is launched by accelerating it along a \(200.0-\mathrm{m}\) incline at 1.25 \(\mathrm{m} / \mathrm{s}^{2}\) starting from rest at point \(A\) (Figure \(3.40 . )\) The incline rises at \(35.0^{\circ}\) above the horizontal,and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point \(A .\)

\(\cdot\) A coyote chasing a rabbit is moving 8.00 \(\mathrm{m} / \mathrm{s}\) due east at one moment and 8.80 \(\mathrm{m} / \mathrm{s}\) due south 4.00 s later. Find (a) the \(x\) and \(y\) components of the coyote's average acceleration during that time and (b) the magnitude and direction of the coyote's average acceleration during that time.
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