Projectile Motion - vertical displacement at specific time.

[wheels94]

Homework Statement

A kicker wishes to kick a field goal, so he sets the ball on the ground 30 yards (27m) from the goal. Assuming he kicks the ball at an angle of 45 degrees and a speed of 27m/s, what is the height of the ball above the ground when it goes over the crossbar.

Known:
R = 27m
Vo = 27ms
theta = 45

Unknown:
t:
Vox:
Voy:
Sy:

Homework Equations

Vox = Vo *cos(theta)
Voy = Vo *sin(theta)
R = Vox*t
Sy = Voy*t+.5*a*(t)^2

The Attempt at a Solution

Step 1: Find Vox:

Vox = Vo *cos(theta)
Vox = 27 *cos(45)
Vox = 19.1m/s

Step 2: Find t:

R = Vox*t
27 = 19.1*t
t = 27/19.1
t = 1.4s

Step 3: Find Voy:

Voy = Vo *sin(theta)
Voy = 27 *sin(45)
Voy = 19.1m/s

Step 4: Find Sy:

Sy = Voy*t+.5*a*(t)^2
Sy = 19.1*1.4 + .5(-9.8)(1.4)^2
Sy = 26.7-9.6
Sy = 17.1m

However, this is not the answer in the back of this book (Practice Makes Perfect Physics by Connie J Wells). The answers given are t=1.4s and h (which I've so far assumed is Sy)= 17.7m

I've only recently started brushing up on my physics over the last year as prep for University, so I'm fully expecting there to be an error here somewhere, but I've already come across one error in this book, is this possibly another one?

 

[gneill]

Your method and calculations look fine. The dual units specified for the distance leads me to suspect that the author originally did the problem in one set of units and simply converted a few values to make it "politically correct" for a metric audience; That's just sloppy and lazy in my opinion.

Redo your calculations with the distance being 30 yards, the speed 90 ft/sec and the acceleration due to gravity 32 ft/sec². The resulting height calculation should convert very closely to 17.7 m.

 

[wheels94]

Your method and calculations look fine. The dual units specified for the distance leads me to suspect that the author originally did the problem in one set of units and simply converted a few values to make it "politically correct" for a metric audience; That's just sloppy and lazy in my opinion.

Redo your calculations with the distance being 30 yards, the speed 90 ft/sec and the acceleration due to gravity 32 ft/sec². The resulting height calculation should convert very closely to 17.7 m.

Comes out at 17.6m, which seems to be as close as I'll be getting on this one. Annoying that the book has some screwy answers, but as long as it's teaching me the correct method it's still a win. Thanks a lot for the help! :)

 

[haruspex]

Hi @wheels94,
There is another useful lesson or two here.
Because of the angle and numbers chosen, the final step in the height calculation is numerically s_x-g. The conversions provided have a 2% error (should be 27.5, not 27). The conversion used for g (32.2 to 9.8) is rather more accurate. The subtraction is between two numbers in approximately the ratio 3:1, so the 2% error becomes a 3% error.
In your own calculation you found time and rounded it to 1.4 s, a 1% error, and used that in the next calculation. This led to an error accumulation, which again was magnified by the subtraction step.
In multistage calculations, it is better to keep everything algebraic through the transition. I.e, don't use the 1.4 as input to the next stage, just use the algebraic expression it came from. Often you will find there is cancellation (there certainly is here).
If you have to carry a numeric value through, preserve more digits than you quote for the intermediate answer. So here, you would answer the first part as 1.4 s, but use 1.414 as input to the next part.