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Chapter 3: Problem 56

A ball is thrown straight upward and returns to the thrower's hand after \(3.00 \mathrm{~s}\) in the air. \(\mathrm{A}\) second ball is thrown at an angle of \(30.0^{\circ}\) with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?

Short Answer

Expert verified
The second ball must be thrown with an approximate speed of 29.43 m/s to reach the same height as the first ball.

Step by step solution

01

Find the initial speed of the first ball

Using the formulas of motion for a projectile under gravity, the initial speed \(v_1\) of the first ball can be found using the equation \(h = v_1 t - 0.5 g t^2\), where \(h\) is the maximum height, \(g\) is the acceleration due to gravity, and \(t\) is the total time of flight. Since the ball returns to its original position, \(h\) is zero, and we have \(0 = v_1 t - 0.5 g t^2\). Solving for \(v_1\), we find \(v_1 = 0.5 g t\). Substituting \(g = 9.81 \, m/s^2\) and \(t = 3.00 \, s\), we find \(v_1 = 14.715 \, m/s\).
02

Find the initial speed of the second ball

For the second ball thrown at an angle, the vertical component of the initial speed \(v_{2y}\) is responsible for the height it reaches. We want the second ball's height to be equal to the first ball's height, which means \(v_{2y} = v_1\). As the second ball is thrown with an angle 胃, we use the trigonometric relation \(v_{2y} = v_2 \sin{\theta}\), where \(v_2\) is the magnitude of the velocity of the second ball, and \(\theta\) is the angle with the horizontal. Solving for \(v_2\), we find \(v_2 = v_2y/\sin{\theta}\). Substituting \(v_2y = v_1 = 14.715 m/s\) and \(\theta = 30^{\circ} = \pi/6 \) radians, we find \(v_2 \approx 29.43 \, m/s\).
03

Conclusion

In order for the second ball to reach the same height as the first ball, it must be thrown with an initial speed of approximately 29.43 m/s.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. When we analyze projectile motion, we often use kinematic equations to describe the trajectory, speed, and acceleration of the projectile at various points.
  • The fundamental kinematic equation for vertical motion under gravity is: \( h = v_i t - \frac{1}{2} g t^2 \).
  • Here, \( h \) is the height, \( v_i \) is the initial velocity, \( t \) is the time, and \( g \) is the acceleration due to gravity, typically \( 9.81 \ m/s^2 \).
When a ball is thrown vertically upwards, it momentarily comes to rest at its highest point before falling back down. The total time in the air includes the time to reach the top and back to the start.
In this scenario, we can find the initial speed by setting the maximum height \( h \) to zero since the ball returns to the same height after its flight. Using this principle, we solve for the initial speed necessary for the ball to reach its peak height before returning.
Projectile Range
Projectile range refers to the horizontal distance a projectile travels during its motion. However, in our scenario, we focus on achieving the same maximum height rather than range. Understanding projectile range concepts aids in determining the necessary launch speed and angle for different motion trajectories.
  • The range of a projectile is maximized when it's launched at a 45-degree angle, given a fixed initial speed.
  • Horizontal and vertical components of motion are independent, meaning the horizontal velocity does not influence the vertical position directly.
For the second ball thrown at a 30-degree angle, its vertical motion, and therefore its range, are dictated by the vertical component of its initial velocity. The goal is to match the height reached by the vertically thrown ball, which solely depends on its initial vertical speed.
To achieve this, the vertical speed component \( v_{2y} \) must equal the initial speed of the first ball, ensuring the same peak height before descending.
Trigonometry
Trigonometry involves studying relationships between the angles and sides of triangles. It's crucial in projectile motion to resolve a projectile's initial velocity into its horizontal and vertical components.
  • In our exercise, the vertical component \( v_{2y} \) is found using the sine of the angle: \( v_{2y} = v_2 \sin{\theta} \).
  • The angle \( \theta \) given is \( 30^{\circ} \), and using known trigonometric values, \( \sin{30^{\circ}} = 0.5 \).
By employing trigonometry, we calculate the required initial speed \( v_2 \) that ensures the ball reaches the same height as the first ball. Thus, the vertical component aligns with the initial vertical velocity derived from the kinematic equations for a ball thrown straight up.
In practice, learning to decompose vectors into components using trigonometry helps in understanding the displacement and trajectory of various projectiles in physics problems.

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Most popular questions from this chapter

A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance \(x\) away from the gun. The pellet hits the target a vertical distance \(y\) below the gun. (a) Show that the position of the pellet when traveling through the air is given by \(y=A x^{2}\), where \(A\) is a constant. (b) Express the constant \(A\) in terms of the initial (muzzle) velocity and the free-fall acceleration. (c) If \(x=3.00 \mathrm{~m}\) and \(y=0.210 \mathrm{~m}\), what is the initial speed of the pellet?

If a person can jump a maximum horizontal distance (by using a 45 projection angle) of \(3.0 \mathrm{~m}\) on Earth, what would be his maximum range on the Moon, where the free-fall acceleration is \(g / 6\) and \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\) ? Repeat for Mars, where the acceleration due to gravity is \(0.38 g .\)

Vector \(\mathrm{A}\) has a magnitude of 29 units and points in the positive \(y\) -direction. When vector \(\overrightarrow{\mathbf{B}}\) is added to \(\overrightarrow{\mathbf{A}}\), the resultant vector \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathbf{B}}\) points in the negative \(y\) -direction with a magnitude of 14 units. Find the magnitude and direction of \(\overrightarrow{\mathbf{B}}\).

An artillery shell is fired with an initial velocity of \(300 \mathrm{~m} / \mathrm{s}\) at \(55.0^{\circ}\) above the horizontal. To clear an avalanche, it explodes on a mountainside \(42.0 \mathrm{~s}\) after firing. What are the \(x\) -and \(y\) -coordinates of the shell where it explodes, relative to its firing point?

A quarterback throws a football toward a receiver with an initial speed of \(20 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. At that instant the receiver is \(20 \mathrm{~m}\) from the quarterback. In what direction and with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?
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