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MS_SUMTER2010 · Feb 23, 2010, 01:36 PM

A 1.6 m tall girl throws a football at an angle of 41 degrees from the horizontal and at an initial velocity of 9.40 m/s. How far away from the girl will it land?

Choices are:

4.55 m
5.90 m
8.90 m
10.5 m

I say it is 8.90 m

ebaines · Feb 23, 2010, 01:48 PM

I don't agree. How did you get your answer?

MS_SUMTER2010 · Feb 24, 2010, 01:56 PM

My teacher says it is 10.5 m but I don't know how he got that

ebaines · Feb 24, 2010, 02:21 PM

This problem can be broken into two steps:

1. Calculate how long it takes for the ball to hit the ground, then
2. Determine the horizontal distance the ball travels during that time.

A key concept to remember is that you treat the vertical motion of the ball as independent from the horizonatl motion. The vertical motion is driven by (a) the initial vertical velocity of the ball, and (b) the effect of gravity, which slows the upward rise of the ball, and eventually causes it to fall to the ground. The equation of motion is:

$ y(t) = y_0 + v_0*t + \frac 1 2 g t^2 $

where $y_0$ is the initial displacement, $v_0$ is the initial velocity in the vertical direction, and $a$ is the acceleration, which in this case is -g, or $-9.8 m/s^2$ . The initial velocity in the vertical direction is found by multiplying the initial velocity (9.4 m/s ) by the sin of 41 degrees. So you have:

$ y(t) = 1.6m + 9.4 \frac m s * sin(41 ^{\small o})*t - \frac 1 2 * 9.8 \frac m {s ^2} * t^2. $

You need to find the value of t when y(t) = 0 ; that is, when the ball hits the ground. This is a quadratic equation which has one positive value of t. So that's how many seconds it takes to hit the ground.

The next step is to calculate how far the ball moves horizontaly during time t. The velocity in the horizontal directoin is constant, since there is no force applied in the horizontal direction to make it speeed up or slow down. So the equation of motion is:

$ x(t) = x_0 + v*t \\ x(t) = 0 m + 9.4 \frac m s * cos(41 ^{\small o}) * t $

Plug the value of t that you got from the first step, and that will give you the horizontal distance the ball travels. You should find the answer to be 10.5 m. Post back if you get stuck on anything.

MS_SUMTER2010 · Feb 24, 2010, 02:42 PM

I GOT 1.6 for the first part. Is that right?

ebaines · Feb 24, 2010, 02:50 PM

I get 1.48 sec. Are you using 9.8 m/s as the value of g?

MS_SUMTER2010 · Feb 24, 2010, 02:53 PM

Yes

MS_SUMTER2010 · Feb 24, 2010, 02:57 PM

I put in this:

y(t)= (1.60m)+(9.4m/s)sin(41)*(0)+-1/2(9.8)(0 squared)

ebaines · Feb 24, 2010, 03:02 PM

You found the at time t = 0. Since the girl is 1.6 meters tall, of course you get y(0) = 1.6 m.

What you want to determine is when does y(t) = 0 , because the ground is at height 0 meters. You should solve this for t:

0 = 1.6 + 9.4 sin(41) * t - 1/2 (9.8) t^2

Like I said earlier - you'll need to use the quadratic formula to find t.

MS_SUMTER2010 · Feb 24, 2010, 03:08 PM

what are the values for a b n c?

ebaines · Feb 24, 2010, 03:24 PM

Originally Posted by MS_SUMTER2010

what are the values for a b n c?

I gave it to you:

0 = 1.6 + 9.4 sin(41) * t - 1/2 (9.8) t^2

MS_SUMTER2010 · Feb 24, 2010, 03:33 PM

OK I got it now! Thanks.. I hate physics as you can see because I never know how to work out the problems.