- $33.3 \textrm{ rad/s}$
- $500 \textrm{ N}$
- $40.8 \textrm{ m}$

### Solution video

# OpenStax College Physics, Chapter 6, Problem 8 (Problems & Exercises)

### Calculator Screenshots

*ma*and then it'd be mass times change in velocity over time. That's essentially what we have here, mass, if you factor it out, it'd be multiplied by change in velocity divided by time. The initial velocity is zero, and so we just say

*mv*final divided by time. That's point five kilograms mass of the football times 20 meters per second, its final velocity, divided by 20 milliseconds, hich we have to rate as times 10 to the minus 3 seconds, and we get 500 newtons. The next question is, what is the maximum possible range of the football? Range refers to what horizontal distance could it possibly cover? The football starts here and then it's kicked by the person and it'll travel in an arc and it'll get, say, here. The question is, what is this distance, the maximum possible distance? That's our max, this is the range maximum. Now in chapter 3 equation 71, we're given a formula for the range. It's the initial velocity squared divided by acceleration due to gravity, multiplied by sine of 2 times the launch angle. That's the angle with respect to horizontal. We want to know what the maximum ranges though so we don't need to concern ourselves with this angle here. We want to know what maximum value could this possibly be. Now the sine function starts at zero for zero degrees, and it goes up to a maximum of one at 90 degrees, and then it goes back down to zero at 180, and then negative one at 270, and then back to zero again at 360. The important thing to notice is that the maximum possible value for the sine function is 1. We can work out that theta should be 45 degrees to get the maximum range. But that's beside the point, the point is that the maximum that this whole thing could possibly be is 1. We can replace it with a number 1. The maximum range then is the initial velocity squared divided by

*g*. That's 20 meters per second squared divided by 9.8 meters per second squared, and that is 40.8 meters is the maximum possible range, assuming there's no air friction.

## Comments

Can you explain to me why t is written as 20.0 E -3 on part b?

Hi alexvaldez-west, thank you for the comment. The question says the person's foot is in contact with the ball for 20.0 ms, which has units *milli*seconds. The prefix *milli* means *multiply by 10 to the power of negative 3*. "E -3" is how $\times 10^{-3}$ is written on many calculators.

Hope this helps, and all the best with your studies,

Shaun

In part b you said the initial velocity was 0 but then in part c to find the range you said the initial velocity was 20. Why did it change?

Hi c-bachman, thank you for the question. In part (b) the ball is being kicked. Yes, it starts at rest, but at the end of the contact time with the foot the ball is accelerated to 20 m/s according to the question. In part (c) we're meant to find the range given that the ball has finished being kicked, after 20 milliseconds, to a speed of 20 m/s. 20m/s is the ball's initial launch speed and we make an assumption about the angle such that the range is at its maximum.

Hope that helps,

Shaun