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…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter2: Motion In One Dimension

Section: Chapter Questions

Problem 49P

See similar textbooks

Related questions

Concept explainers

Question

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.

Step by step

Solved in 4 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions