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Chapter 5: Problem 11

A projectile has a time of flight \(T\) and range \(R\). If the time of flight is doubled, keeping the angle of projection same what happens to the range? (1) \(R / 4\) (2) \(\mathrm{R} / 2\) (3) \(2 \mathrm{R}\) (4) \(4 \mathrm{R}\)

Short Answer

Expert verified
The range becomes \(4R\), so the correct answer is (4).

Step by step solution

01

Initial Formula for Range and Time

The range \(R\) of a projectile launched at an angle \(\theta\) with initial velocity \(v\) is given by \[ R = \frac{v^2 \sin(2\theta)}{g} \]The time of flight \(T\) is given by\[ T = \frac{2v \sin(\theta)}{g} \]
02

Expression of Range in Terms of Time

Re-express the range \(R\) using time \(T\). Rewrite the initial velocity \(v\) in terms of time \[ v = \frac{gT}{2\sin(\theta)} \]Substitute for \(v\) in the range formula:\[ R = \frac{(gT/(2\sin(\theta)))^2 \sin(2\theta)}{g} \]Simplify the equation:\[ R = \frac{g^2 T^2 \sin(2\theta)}{4g \sin^2(\theta)} = \frac{g T^2 \sin(2\theta)}{4 \sin^2(\theta)} \]
03

Analyze the Effect of Doubling the Time

Given that the time of flight \(T\) is doubled, we take the new time of flight as \(T' = 2T\). Substitute \(T'\) in the range equation:\[ R' = \frac{g (2T)^2 \sin(2\theta)}{4 \sin^2(\theta)} \]This becomes:\[ R' = \frac{g \cdot 4T^2 \sin(2\theta)}{4 \sin^2(\theta)} = \frac{4g T^2 \sin(2\theta)}{4 \sin^2(\theta)} \]Simplify this to:\[ R' = 4 R \]

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Key Concepts

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

Time of Flight
The time of flight is the total time a projectile stays in the air. It depends on two main factors:
  • Initial velocity
  • Angle of projection
When a projectile is launched at a certain angle \(\theta\) with an initial speed \(v\), the time of flight \(T\) is given by the formula:\[T = \frac{2v \sin(\theta)}{g}\]In this formula, \(g\) represents the acceleration due to gravity, which is approximately 9.8 m/s虏 on Earth's surface. This equation shows that the time of flight increases with a greater initial speed and a steeper angle of projection.
When we double the time of flight while keeping the angle constant, it implies that somehow the initial speed or the conditions affecting it have changed to make the projectile stay longer in the air.
Range of a Projectile
The range of a projectile is how far it travels horizontally during its flight. This measurement also depends on the initial velocity and the angle of projection. The formula for calculating the range \(R\) is:\[R = \frac{v^2 \sin(2\theta)}{g}\]This expression involves the sine of twice the angle, revealing that the range is maximized when the projectile is launched at an angle of 45 degrees.
If we consider our exercise and keep the angle constant while doubling the time of flight \(T\), we find that the range becomes four times greater. Specifically, when the time of flight \(T\) is doubled, substituting \(2T\) into the range formula results in \(4R\).
Thus, the range increases significantly, showing how interconnected these variables are.
Angle of Projection
The angle of projection is a critical factor that influences both the time of flight and the range of a projectile. It dictates the trajectory's shape and maximum reach. As seen in the formulas:
  • The time of flight \(T\) includes the term \(\sin(\theta)\)
  • The range \(R\) includes \(\sin(2\theta)\)
These trigonometric functions highlight how the angle affects the motion outcome. While a 45-degree angle maximizes the range, the time of flight benefits from larger angles, leading to a higher vertical reach, hence a longer airtime.
Understanding the role of the projection angle helps predict how changes like doubling the time of flight alter the range. It explains why precise adjustments to this angle can optimize a projectile's distance and airtime.

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Most popular questions from this chapter

We know that when a boat travels in water, its net velocity w.r.t. ground is the vector sum of two velocities. First is the velocity of boat itself in river and other is the velocity of water w.r.t. ground. Mathematically: \(\vec{v}_{\text {boat }}=\vec{v}_{\text {boat, water }}+\vec{v}_{\text {water }}\) Now given that velocity of water w.r.t. ground in a river is \(u\). Width of the river is \(d\). A boat starting from rest aims perpendicular to the river with an acceleration of \(a=5 t\), where \(t\) is time. The boat starts from point \((1,0)\) of the coordinate system as shown in figure. Assume SI units. Obtain the total time taken to cross the river. (1) \((3 d / 5)^{1 / 3}\) (2) \((6 d / 5)^{1 / 3}\) (3) \((6 d / 5)^{1 / 2}\) (4) \((2 d / 3)^{1 / 3}\)

A projectile is fired from level ground at an angle \(\theta_{a b_{b_{1}}}\) the horizontal. The elevation angle \(\phi\) of the highest \(p_{0 i n}\). seen from the launch point is related to \(\theta\) by the relation (1) \(\tan \phi=2 \tan \theta\) (2) \(\tan \phi=\tan \theta\) (3) \(\tan \phi=\frac{1}{2} \tan \theta\) (4) \(\tan \phi=\frac{1}{4} \tan \theta\)

A river is flowing towards with a velocity of \(5 \mathrm{~m} \mathrm{~s}^{-1}\). The boat velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\). The boat crosses the river by shortest path. Hence, (1) The direction of boat's velocity is \(30^{\circ}\) west of north. (2) The direction of boat's velocity is north-west. (3) Resultant velocity is \(5 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\). (4) Resultant velocity of boat is \(5 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\).

A river flows with a speed more than the maximum speed with which a person can swim in still water. He intends to cross the river by the shortest possible path (i.e., he wants to reach the point on the opposite bank which directly opposite to the starting point). Which of the following is correct? (1) He should start normal to the river bank. (2) He should start in such a way that he moves normal to the bank, relative to the bank. (3) He should start in a particular (calculated) direction making an obtuse angle with the direction of water current. (4) The man cannot cross the river in that way.

A body is thrown with the velocity \(v_{0}\) at an angle of \(\theta\) to the horizon. Determine \(v_{0}\) in \(\mathrm{ms}^{-1}\) if the maximum height attained by the body is \(5 \mathrm{~m}\) and at the highest point of its trajectory the radius of curvature is \(r=3 \mathrm{~m}\). Neglect air resistance. [Use \(\sqrt{80}\) as 9 ]
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