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Chapter 3: Problem 72

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 ball would reach a maximum height of 12.0 meters if thrown straight upward.

Step by step solution

01

Understand the Problem

We are given that a ball is thrown at an angle of \(52^\circ\) with an initial speed \(v_0\) and reaches a maximum height of 7.5 meters. We need to find the maximum height if it were thrown vertically with the same initial speed.
02

Decompose Initial Velocity

When the ball is thrown at an angle, its initial velocity \(v_0\) can be divided into two components: the vertical component \(v_{0y} = v_0 \sin 52^\circ\) and the horizontal component \(v_{0x} = v_0 \cos 52^\circ\). For the maximum height calculation, we are interested in the vertical component.
03

Use Kinematic Equation

The kinematic equation for maximum height \(h\) of an object thrown vertically is given by: \(h = \frac{v_{0y}^2}{2g}\), where \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\)).
04

Calculate Vertical Velocity Component

Since the ball reaches a maximum height of 7.5 meters when thrown at 52 degrees, substitute into the height equation: \(7.5 = \frac{(v_0 \sin 52^\circ)^2}{2 \times 9.81}\). Solve this equation for \(v_0 \sin 52^\circ\).
05

Find Initial Speed \(v_0\sin 52^\circ\)

Rearrange the equation to get \(v_0 \sin 52^\circ = \sqrt{7.5 \times 2 \times 9.81}\) and calculate it. This gives us the magnitude of the vertical component of the initial velocity required to reach the maximum height of 7.5 meters.
06

Calculate Maximum Height for Vertical Throw

Given that \(v_0 \sin 52^\circ\) represents the vertical component of \(v_0\), if the ball is thrown directly upward with speed \(v_0\), it will have the whole speed \(v_0\) as vertical (i.e., \(v_{0y} = v_0\)). Use the equation \(h = \frac{v_{0}^2}{2g}\) to find the new maximum height. Substitute the value of \(v_0\) and solve.

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Key Concepts

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

Kinematic Equations
Kinematic equations play a crucial role in analyzing projectile motion. In our exercise, they help us determine how far and how high an object moves through the air. These equations describe the motion of objects under constant acceleration, like gravity. This allows us to calculate various parameters such as displacement, velocity, and time.
For finding maximum height, we use the equation:\[h = \frac{v_{0y}^2}{2g}\]Here:
  • \(v_{0y}\) is the initial vertical velocity,
  • and \(g\) is the acceleration due to gravity.
When you understand these equations, you can determine how changes in initial conditions affect the motion.
Vertical Motion
Vertical motion refers to the upward or downward movement of an object thrown into the air. In projectile motion, gravity impacts how high the object goes. As soon as the ball leaves hand, it starts losing speed due to gravity, eventually coming to a stop at the peak height before falling back down.
This motion is influenced by:
  • Gravity, which accelerates the object downward at \(9.81 \text{ m/s}^2\),
  • and the initial vertical velocity, which is influenced by the angle and speed of the throw.
At the maximum height, the vertical component of the object's velocity becomes zero. Hence, using the kinematic equation, we can determine how changing initial velocities, such as throwing the ball straight upward versus at an angle, influences that height.
Initial Velocity Components
When dealing with projectile motion, breaking the velocity into components is key. Initial velocity can be broken down into two components: vertical and horizontal. These components determine how the object will move through the air.
In our problem, a ball is thrown at an angle of 52 degrees, which means:
  • Vertical component \((v_{0y}) = v_0 \sin 52^\circ\)
  • Horizontal component \((v_{0x}) = v_0 \cos 52^\circ\)
To find the height reached by the ball thrown vertically:
  • We focus on the vertical component because it tells us how high the ball will go when gravity is the only force acting on it.
  • If thrown directly upward with speed \(v_0\), all speed assists in climbing vertically, maximizing the height.
Understanding these components can provide insight into how different angles affect the projectile's motion and ensure more accurate predictions in real-world applications.

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Most popular questions from this chapter

An airplane with a speed of \(97.5 \mathrm{m} / \mathrm{s}\) is climbing upward at an angle of \(50.0^{\circ}\) with respect to the horizontal. When the plane's altitude is \(732 \mathrm{m},\) the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

A swimmer, capable of swimming at a speed of \(1.4 \mathrm{m} / \mathrm{s}\) in still water (i.e., the swimmer can swim with a speed of \(1.4 \mathrm{m} / \mathrm{s}\) relative to the water), starts to swim directly across a \(2.8-\mathrm{km}\) -wide river. However, the current is \(0.91 \mathrm{m} / \mathrm{s},\) and it carries the swimmer downstream. (a) How long does it take the swimmer to cross the river? (b) How far downstream will the swimmer be upon reaching the other side of the river?

Two friends, Barbara and Neil, are out rollerblading. With respect to the ground, Barbara is skating due south at a speed of \(4.0 \mathrm{m} / \mathrm{s} .\) Neil is in front of her. With respect to the ground, Neil is skating due west at a speed of \(3.2 \mathrm{m} / \mathrm{s} .\) Find Neil's velocity (magnitude and direction relative to due west), as seen by Barbara.

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 \(D\).

A golf ball rolls off a horizontal cliff with an initial speed of \(11.4 \mathrm{m} / \mathrm{s} .\) The ball falls a vertical distance of \(15.5 \mathrm{m}\) into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed \(v\) of the ball just before it strikes the water?
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