You threw a tennis ball from a balcony 3.3 m above the ground upward at 40 degrees above horizontal with an initial speed of 4.5 m/s. How long was the tennis in the air?
You threw a tennis ball from a balcony 3.3 m above the ground upward at 40 degrees above horizontal with an initial speed of 4.5 m/s. How long was the tennis in the air?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Projectile Motion Problem**
_Read the following problem and use the concepts of projectile motion to solve it._
**Problem Statement:**
You threw a tennis ball from a balcony 3.3 m above the ground upward at 40 degrees above horizontal with an initial speed of 4.5 m/s. How long was the tennis ball in the air?
**Explanation:**
To solve for the time the tennis ball was in the air, you need to decompose the motion into horizontal and vertical components and apply the kinematic equations.
First, break the initial velocity into horizontal and vertical components:
- \( v_{0x} = v_0 \cdot \cos(\theta) \)
- \( v_{0y} = v_0 \cdot \sin(\theta) \)
where:
- \( v_0 \) = 4.5 m/s (initial speed)
- \( \theta \) = 40 degrees (angle above horizontal)
Next, use the vertical motion to determine the time in the air:
- The vertical displacement (Δy) is -3.3 m (since the ball is thrown upwards and lands below its original height).
- The vertical component of the initial velocity \( v_{0y} = 4.5 \cdot \sin(40^\circ) \)
- The acceleration due to gravity \( g \) is approximately -9.8 m/s².
From the kinematic equation:
\[ \Delta y = v_{0y} \cdot t + \frac{1}{2} a t^2 \]
Substitute the known values:
\[ -3.3 m = (4.5 \cdot \sin(40^\circ)) \cdot t + \frac{1}{2} (-9.8) t^2 \]
Solve this quadratic equation for \( t \), which gives you the total time the tennis ball is in the air.
Understanding the components and following through with the kinematic equations will yield the desired time. Using these principles can help solve similar problems in projectile motion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e60ae64-4458-4b89-bb0e-7dd9dd5dc5fe%2Fc545be7f-cc8a-4a61-909f-39ddf000c6ed%2Fz4gi65q_processed.png&w=3840&q=75)
Transcribed Image Text:**Projectile Motion Problem**
_Read the following problem and use the concepts of projectile motion to solve it._
**Problem Statement:**
You threw a tennis ball from a balcony 3.3 m above the ground upward at 40 degrees above horizontal with an initial speed of 4.5 m/s. How long was the tennis ball in the air?
**Explanation:**
To solve for the time the tennis ball was in the air, you need to decompose the motion into horizontal and vertical components and apply the kinematic equations.
First, break the initial velocity into horizontal and vertical components:
- \( v_{0x} = v_0 \cdot \cos(\theta) \)
- \( v_{0y} = v_0 \cdot \sin(\theta) \)
where:
- \( v_0 \) = 4.5 m/s (initial speed)
- \( \theta \) = 40 degrees (angle above horizontal)
Next, use the vertical motion to determine the time in the air:
- The vertical displacement (Δy) is -3.3 m (since the ball is thrown upwards and lands below its original height).
- The vertical component of the initial velocity \( v_{0y} = 4.5 \cdot \sin(40^\circ) \)
- The acceleration due to gravity \( g \) is approximately -9.8 m/s².
From the kinematic equation:
\[ \Delta y = v_{0y} \cdot t + \frac{1}{2} a t^2 \]
Substitute the known values:
\[ -3.3 m = (4.5 \cdot \sin(40^\circ)) \cdot t + \frac{1}{2} (-9.8) t^2 \]
Solve this quadratic equation for \( t \), which gives you the total time the tennis ball is in the air.
Understanding the components and following through with the kinematic equations will yield the desired time. Using these principles can help solve similar problems in projectile motion.
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