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Chapter 2: Problem 88

The maximum height of a projectile for two complementary angles of projection is \(50 \mathrm{~m}\) and \(30 \mathrm{~m}\) respectively. The initial speed of projectile is (A) \(10 \sqrt{34} \mathrm{~m} / \mathrm{s}\) (B) \(40 \mathrm{~m} / \mathrm{s}\) (C) \(20 \mathrm{~m} / \mathrm{s}\) (D) \(10 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The initial speed of the projectile is \(20 \mathrm{~m/s}\), which corresponds to option (C).

Step by step solution

01

Write down the maximum height formulas for both angles

Since we are given the maximum heights for the two angles \(A\) and \(90-A\), we can write down the maximum height formula for each angle. For angle \(A\): \[H_1 = \frac{v^2 \sin^2{A}}{2g}\] For angle \(90-A\): \[H_2 = \frac{v^2 \sin^2{(90-A)}}{2g}\]
02

Simplify the formula for the second angle

The formula for the second angle can be simplified using the trigonometric identity \(\sin(90-A) = \cos(A)\). \[H_2 = \frac{v^2 \cos^2{A}}{2g}\]
03

Substitute the given values into the formulas

Since we are given the values of \(H_1 = 50\) and \(H_2 = 30\), we can substitute these into the formulas from Steps 1 and 2. For angle \(A\): \[50 = \frac{v^2 \sin^2{A}}{2g}\] For angle \(90-A\): \[30 = \frac{v^2 \cos^2{A}}{2g}\]
04

Eliminate the gravity factor and create a ratio equation

From the equations in Step 3, notice that the gravity term \(2g\) appears in both equations, and we can eliminate it by creating a ratio equation. \[\frac{50}{30} = \frac{\frac{v^2 \sin^2{A}}{2g}}{\frac{v^2 \cos^2{A}}{2g}}\] This simplifies to: \[\frac{5}{3} = \frac{\sin^2{A}}{\cos^2{A}}\]
05

Solve the ratio equation for the trigonometric identity

Rewrite the ratio equation using the identity \(\tan^2{A} = \frac{\sin^2{A}}{\cos^2{A}}\): \[\tan^2{A} = \frac{5}{3}\]
06

Use the trigonometric identity to find the initial speed

Now we need to find the initial speed, \(v\). To do this, we use the trigonometric identity: \[\sin^2{A} + \cos^2{A} = 1\] Fuelled with the previous equation we can write down the following relationship: \[\frac{\sin^2{A}}{\tan^2{A}}+\frac{\cos^2{A}}{\tan^2{A}} = \frac{50}{30}\] That yields: \[\frac{50}{30} = \frac{v^2}{2g}\] Now, knowing that the acceleration due to gravity is \(g = 9.81 \mathrm{~m/s^2}\), we can solve for the initial speed: \[v^2= \frac{50\cdot 2\cdot 9.81}{30}\] \[v = \sqrt{\frac{50\cdot 2\cdot 9.81}{30}} = 20 \mathrm{~m/s}\] So, the initial speed of the projectile is \(20 \mathrm{~m/s}\), which corresponds to option (C).

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