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A ball is thrown upward at a speed \(v_{0}\) at an angle of \(52^{\circ}\) above the horizontal. It reaches a maximum height of \(7.5 \mathrm{m}\). How high would this ball go if it were thrown straight upward at speed \(v_{0} ?\)

Short Answer

Expert verified
The maximum height when thrown straight upward is higher than 7.5 m.

Step by step solution

01

Determine Initial Speed in Vertical Projection

When a ball is thrown at an angle, its initial velocity can be split into a horizontal and vertical component. For a projection angle of \(52^{\circ}\) and maximum height of \(7.5 \mathrm{m}\), we first find the vertical component \(v_{0y}\) of the initial velocity, which affects the maximum height. The maximum height \(H\) can be calculated using the formula \(H = \frac{v_{0y}^2}{2g}\), where \(g\) is the acceleration due to gravity \(9.8 \mathrm{m/s^2}\). Rearranging gives: \(v_{0y} = \sqrt{2gH}\). Plug in \(H = 7.5\mathrm{m}\) to get \(v_{0y}.\)
02

Calculate Total Initial Speed

In the given problem, the relationship between the vertical component \(v_{0y}\) and the initial speed \(v_0\) is established through trigonometry as \(v_{0y} = v_{0} \sin(52^{\circ})\). To find \(v_{0}\), use the previously calculated \(v_{0y}\) and rearrange the expression: \(v_{0} = \frac{v_{0y}}{\sin(52^{\circ})}\). Substitute the value of \(v_{0y}\) obtained in Step 1 to compute \(v_{0}.\)
03

Evaluate Maximum Height for Vertical Throw

If the ball were thrown straight upward, its entire initial velocity \(v_{0}\) contributes to reaching the maximum height. The new maximum height \(H'\) is calculated using \(H' = \frac{v_{0}^2}{2g}\). Substitute the calculated value of \(v_{0}\) from Step 2 into this expression to find the new maximum height \(H'\).

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Key Concepts

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

Vertical Component of Velocity
When an object is thrown at an angle, its velocity can be broken down into two components: horizontal and vertical. The vertical component of velocity is crucial for predicting how high and how long an object will travel vertically.

For a given initial velocity, the vertical component, denoted as \( v_{0y} \), is determined using trigonometry. If \( v_0 \) is the initial speed and \( \theta \) the angle of projection, then the vertical component is given by \( v_{0y} = v_0 \sin(\theta) \).

In our case, for an angle of \(52^{\circ}\) and known vertical height, we apply the trigonometric function to find the portion of the initial velocity that acts in the vertical direction. This vertical component is directly related to the object's ascent against gravity.
Maximum Height Calculation
To determine the maximum height an object reaches, we use the vertical component of its initial velocity. This is because only the vertical motion defines how high the object will go.

The formula for calculating the maximum height \(H\) reached by a projectile is \(H = \frac{v_{0y}^2}{2g}\), where \(g\) is the acceleration due to gravity, \(9.8 \, \mathrm{m/s^2}\). It shows how the square of the vertical component of velocity contributes to the height against the constant pull of gravity.

In scenarios where the same object is thrown directly upwards rather than at an angle, the entire initial speed \(v_0\) is used for the height calculation. Thus, the maximum height in such a direct throw is given by \(H' = \frac{v_{0}^2}{2g}\). This means, for the direct upward throw, all the energy aids in overcoming gravity, potentially reaching a higher point.
Kinematics in Two Dimensions
Projectile motion is quintessentially a two-dimensional kinematic problem involving both horizontal and vertical components of motion. Motion in these directions occurs simultaneously but operates independently of each other.

The initial speed of a projectile can be split into two parts: horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) velocities. Kinematics in two dimensions analyzes these using respective projectile motion equations and Newton's laws.

Horizontal motion remains constant due to the absence of horizontal forces (ignoring air resistance), while vertical motion is influenced by gravity. Thus, once the initial split is done via trigonometric functions, later calculations of time of flight, range, and maximum height are carried out independently in these two dimensions. Understanding this allows us to solve complex motion scenarios by simplifying them into more manageable calculations.

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Most popular questions from this chapter

As a tennis ball is struck, it departs from the racket horizontally with a speed of \(28.0 \mathrm{m} / \mathrm{s}\). The ball hits the court at a horizontal distance of \(19.6 \mathrm{m}\) from the racket. How far above the court is the tennis ball when it leaves the racket?

A ferryboat is traveling in a direction \(38.0^{\circ}\) north of east with a speed of \(5.50 \mathrm{m} / \mathrm{s}\) relative to the water. A passenger is walking with a velocity of \(2.50 \mathrm{m} / \mathrm{s}\) due east relative to the boat. What is the velocity (magnitude and direction) of the passenger with respect to the water? Determine the directional angle relative to due east.

A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of \(5.3 \mathrm{m} / \mathrm{s}\), hoping to land on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the two buildings is \(D,\) and the roof of the adjacent building is \(2.0 \mathrm{m}\) below the jumping-off point. Find the maximum value for \(D\).

You are in a hot-air balloon that, relative to the ground, has a velocity of \(6.0 \mathrm{m} / \mathrm{s}\) in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is \(2.0 \mathrm{m} / \mathrm{s}\). What are the magnitude and direction of the hawk's velocity relative to the ground? Express the directional angle relative to due east.

Two cannons are mounted as shown in the drawing and rigged to fire simultaneously. They are used in a circus act in which two clowns serve as human cannonballs. The clowns are fired toward each other and collide at a height of \(1.00 \mathrm{m}\) above the muzzles of the cannons. Clown \(\mathrm{A}\) is launched at a \(75.0^{\circ}\) angle, with a speed of \(9.00 \mathrm{m} / \mathrm{s}\). The horizontal separation between the clowns as they leave the cannons is \(6.00 \mathrm{m}\). Find the launch speed \(v_{0 \mathrm{B}}\) and the launch angle \(\theta_{\mathrm{B}}\left(>45.0^{\circ}\right)\) for clown \(\mathrm{B} .\)

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