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Chapter 4: Problem 45

The maximum range of projectile fired with some initial velocity is found to be \(1000 \mathrm{~m}\), in the absence of wind and air resistance. The maximum height reached by the projectile is [Orissa JEE 2009] (a) \(250 \mathrm{~m}\) (b) \(500 \mathrm{~m}\) (c) \(1000 \mathrm{~m}\) (d) \(2000 \mathrm{~m}\)

Short Answer

Expert verified
(a) 250 m

Step by step solution

01

Identify Given Information

The problem states that the maximum range of a projectile is 1000 meters. We know from projectile physics that the maximum range occurs when the projectile is fired at a 45-degree angle.
02

Recall Range Formula

The formula for the range of a projectile is given by \( R = \frac{v^2 \sin 2\theta}{g} \). For maximum range, \(\theta = 45^\circ\), so the formula simplifies to \( R = \frac{v^2}{g} \).
03

Solve for Velocity

Using \( R = 1000 \) meters and \( g = 9.8 \mathrm{~m/s^2} \), substitute into the range formula: \( 1000 = \frac{v^2}{9.8} \). Solve for \(v^2\): \( v^2 = 1000 \times 9.8 \). Calculate \( v^2 = 9800 \).
04

Find Maximum Height Formula

The formula for the maximum height of a projectile is \( H = \frac{v^2 \sin^2 \theta}{2g} \). For maximum height during maximum range, \(\theta = 45^\circ\), so \( \sin^2 45^\circ = \frac{1}{2} \).
05

Calculate Maximum Height

Substitute \( v^2 = 9800 \) and \( \sin^2 45^\circ = \frac{1}{2} \) into the height formula: \( H = \frac{9800 \times \frac{1}{2}}{2 \times 9.8} \). Simplify to obtain \( H = \frac{4900}{19.6} \). Calculate \( H \approx 250 \mathrm{~m} \).
06

Choose Correct Answer

The calculated maximum height is approximately 250 meters, matching option (a).

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Key Concepts

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

Range of Projectile
Projectile motion occurs when an object is launched into the air and moves under the influence of gravity, following a curved path. The range of a projectile refers to the horizontal distance it travels before landing. To achieve maximum range, a projectile should be launched at a 45-degree angle, where both vertical and horizontal components of the velocity contribute equally.The formula to calculate the range, denoted as \( R \), is given by:\[R = \frac{v^2 \sin 2\theta}{g}\]where:
  • \( v \) is the initial velocity.
  • \( \theta \) is the launch angle.
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \mathrm{~m/s^2} \).
For maximum range, set \( \theta = 45^{\circ} \), making \( \sin 2\theta = 1 \). This simplifies the formula to \( R = \frac{v^2}{g} \). This relationship allows us to calculate the initial speed needed for a desired range, assuming ideal conditions without air resistance.
Maximum Height of Projectile
The maximum height of a projectile is the highest vertical position it reaches in its path. This occurs when its vertical component of velocity is zero, at the peak of its flight. To calculate the maximum height \( H \), we use the formula:\[H = \frac{v^2 \sin^2 \theta}{2g}\]where:
  • \( v \) is the initial velocity.
  • \( \theta \) is the launch angle.
  • \( g \) is the acceleration due to gravity.
For maximum height in the context of maximum range conditions (\( \theta = 45^{\circ} \)), the vertical velocity component is \( v \sin 45^{\circ} \). The maximum height thus depends on the square of this vertical component. Practically, understanding and calculating maximum height help determine how high a projectile can go, which is crucial in various applications, from sports to engineering.
Projectile Angle
The angle at which a projectile is launched greatly affects its trajectory, range, and maximum height. Known as the projectile angle, it determines how the initial velocity is distributed between the horizontal and vertical components.When firing for maximum range, set the projectile angle \( \theta \) to 45 degrees. At this angle, the sine and cosine components of the angle are equal, optimizing the distribution of velocity between height and distance.Here's why the projectile angle is crucial:
  • Smaller Angles: These produce longer horizontal trajectories with minimal height, best for targets close to the ground.
  • Angles Greater Than 45 Degrees: These result in a higher flight path but reduced range.
Knowing how to select the appropriate angle is essential for hitting targets accurately, ensuring projectiles hit their intended mark with effectiveness and efficiency.

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Most popular questions from this chapter

A particle is projected with certain velocity at two different angles of projections with respect to horizontal plane so as to have same range \(R\) on a horizontal plane. If \(t_{1}\) and \(t_{2}\) are the time taken for the two paths, the which one of the following relations is correct? [UP SEE 2008] (a) \(t_{1} t_{2}=\frac{2 R}{g}\) (b) \(t_{1} t_{2}=\frac{R}{g}\) (c) \(t_{1} t_{2}=\frac{g}{2 g}\) (d) \(t_{1} t_{2}=\frac{4 R}{g}\)

Two particles are projected in air with speed \(v_{0}\) at angles \(\theta_{1}\) and \(\theta_{2}\) (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices (a) angle of projection : \(\theta_{1}>\theta_{2}\) (b) time of flight : \(T_{1}>T_{2}\) (c) horizontal range : \(R_{1}>R_{2}\) (d) total energy : \(U_{1}>U_{2}\)

Two projectiles \(A\) and \(B\) thrown with speeds in the ratio \(1: \sqrt{2}\) acquired the same heights. If \(A\) is thrown at an angle of \(45^{\circ}\) with the horizontal, the angle of projection of \(B\) will be (a) \(0^{\circ}\) (b) \(60^{\circ}\) (c) \(30^{\circ}\) (d) \(45^{\circ}\) (e) \(15^{\circ}\)

A body of mass \(m\) is thrown upward at an angle \(\theta\) with the horizontal with velocity \(v\). While rising up the velocity of the mass after \(t\) second will be (a) \(\sqrt{[v \cos \theta)^{2}+(v \cdot \sin \theta)^{2}}\) (b) \(\sqrt{(v \cos \theta-v \sin \theta)^{2}-g t}\) (c) \(\sqrt{v^{2}+g^{2} t^{2}-(2 v \sin \theta) g t}\) (d) \(\sqrt{v^{2}+g^{2} t^{2}-(2 v \cos \theta) g t}\)

A particle \(A\) is projected from the ground with an initial velocity of \(10 \mathrm{~ms}^{-1}\) at an angle of \(60^{\circ}\) with horizontal. From what height \(h\) should an another particle \(B\) be projected horizontal with veloeity \(5 \mathrm{~ms}^{-1}\) go that both the particles collide with velocity \(5 \mathrm{~ms}^{-1}\) so that both the particles collide on the ground at point \(C\) if both are projected simultaneously? \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(10 \mathrm{~m}\) (b) \(30 \mathrm{~m}\) [c) \(15 \mathrm{~m}\) (d) \(25 \mathrm{~m}\)
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