23. A ball is thrown vertically upwards from the ground with an initial velocity of96 feet per second. Use the function s (t) = -16t +vt +s where v is the initialvelocity and s is the initial heighta) After how many seconds does the ball attains its maximum height?b) What is the maximum height the ball attains?c) After how many second does the ball return to the ground?

Answered: 23. A ball is thrown vertically upwards…

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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23. A ball is thrown vertically upwards from the ground with an initial velocity of
96 feet per second. Use the function s (t) = -16t +vt +s where v is the initial
velocity and s is the initial height
a) After how many seconds does the ball attains its maximum height?
b) What is the maximum height the ball attains?
c) After how many second does the ball return to the ground?

Expert Solution

Given data:

The initial velocity, $v = 96 ft / s$

The function for the position of the ball is,

$s (t) = - 16 t^{2} + v t + s$

Solution for (a):

The function for the position of the ball is,

$\begin{array}{rcl} s (t) & = & - 16 t^{2} + v t + s \\ ∴ s (t) & = & - 16 t^{2} + 96 t + s \end{array}$

The velocity of the ball at the maximum height is zero, therefore, differentiate the expression above with respect to time to determine the time required for the ball to reach maximum height,

$\begin{array}{rcl} \frac{d s (t)}{d t} & = & \frac{d}{d t} (- 16 t^{2} + 96 t + s) \\ v (t) & = & - 32 t + 96 \end{array}$

Here, s is constant, therefore, $\frac{d s}{d t} = 0$

Substitute $v (t) = 0$ and solve for t

$\begin{array}{rcl} 0 & = & - 32 t + 96 \\ 32 t & = & 96 \\ t & = & \frac{96}{32} \\ t & = & 3 s \end{array}$

Therefore, the time required for the ball to reach the maximum height is 3 seconds.

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