You throw a ball, starting at a height of 2 meters, of mass 1 kilogram upward with a velocity of v = 24 m/s. Remember that the acceleration due to gravity is g = 9.8. 82 A. Find the position function, s(t) for the ball. -4.9t2 + 24t + 2 B. Approximately how long will the ball be in the air? Round to the nearest thousandth. | 4.98 C. Approximately how high the ball will go? Round to the nearest thousandth.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Can someone please help me with the last part of this question? I thought that I was supposed to use the formula  -b/2a but I am getting the wrong answer. 

Image attached,

Thank You!

**Physics Problem: Projectile Motion**

You throw a ball, starting at a height of 2 meters, of mass 1 kilogram upward with a velocity of \( v = 24 \, \text{m/s} \). Remember that the acceleration due to gravity is \( g = 9.8 \, \frac{\text{m}}{\text{s}^2} \).

A. **Find the position function, \( s(t) \), for the ball.**

\[ s(t) = -4.9t^2 + 24t + 2 \]

B. **Approximately how long will the ball be in the air? Round to the nearest thousandth.**

\[ \text{Time in air} = 4.98 \, \text{seconds} \]

C. **Approximately how high the ball will go? Round to the nearest thousandth.**

\[ \text{Maximum height} = \, \text{(Enter calculation here)} \]
Transcribed Image Text:**Physics Problem: Projectile Motion** You throw a ball, starting at a height of 2 meters, of mass 1 kilogram upward with a velocity of \( v = 24 \, \text{m/s} \). Remember that the acceleration due to gravity is \( g = 9.8 \, \frac{\text{m}}{\text{s}^2} \). A. **Find the position function, \( s(t) \), for the ball.** \[ s(t) = -4.9t^2 + 24t + 2 \] B. **Approximately how long will the ball be in the air? Round to the nearest thousandth.** \[ \text{Time in air} = 4.98 \, \text{seconds} \] C. **Approximately how high the ball will go? Round to the nearest thousandth.** \[ \text{Maximum height} = \, \text{(Enter calculation here)} \]
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