Solve 2 Equations 2 Unknowns: Hibbeler Dynamics 12th Ed.

[Mr Beatnik]

Homework Statement

Ok forgive me as an engineering student but this problem should be easier than it seems. The Problem: It is observed that the skier leaves the ramp A at an angle (Theata=25 degrees) with the horizontal. If he strikes the ground at point B, determine his initial speed,V, and the time of flight,t.

Homework Equations

I have used:

s=vt
s=s+vt+1/2at^2

The dimensions needed are correct and are 100m down the slope alligned, the ramp he leaves from is 4m high and the ramp is angled at a 3,4,5 triangle.

This problem is 12-110 from the Hibbeler Dynamics 12th edition

The Attempt at a Solution

I have setup the equations as

(1) 100(4/5)=Vcos(25)t
(2) -4-100(3/5)=0+Vsin(25)t+(1/2)(-9.81)t^2

I have tried solving (1) for t and plugging in into (2) but come out with a strange decimal and also solving (1) for V and plugging into (2) I can't seem to get it.

I know the answers are supposed to be: V=19.4m/s t=4.54s

Please help!

 

[berkeman]

Homework Statement

Ok forgive me as an engineering student but this problem should be easier than it seems. The Problem: It is observed that the skier leaves the ramp A at an angle (Theata=25 degrees) with the horizontal. If he strikes the ground at point B, determine his initial speed,V, and the time of flight,t.

Homework Equations

I have used:

s=vt
s=s+vt+1/2at^2

The dimensions needed are correct and are 100m down the slope alligned, the ramp he leaves from is 4m high and the ramp is angled at a 3,4,5 triangle.

This problem is 12-110 from the Hibbeler Dynamics 12th edition

The Attempt at a Solution

I have setup the equations as

(1) 100(4/5)=Vcos(25)t
(2) -4-100(3/5)=0+Vsin(25)t+(1/2)(-9.81)t^2

I have tried solving (1) for t and plugging in into (2) but come out with a strange decimal and also solving (1) for V and plugging into (2) I can't seem to get it.

I know the answers are supposed to be: V=19.4m/s t=4.54s

Please help!

A 3-4-5 right triangle does not give you a takeoff angle of 25 degrees.

 

[Mr Beatnik]

Sorry about the confusion and I am aware that a 3-4-5 triangle does not make a 25 degree takeoff. Here is a free body diagram to help describe. The equations are correct btw, I just can't solve the system. Thanks for the help.

 

[berkeman]

Sorry about the confusion and I am aware that a 3-4-5 triangle does not make a 25 degree takeoff. Here is a free body diagram to help describe. The equations are correct btw, I just can't solve the system. Thanks for the help.

Ah, that helps. Where did the "-4" come from?

(2) -4-100(3/5)=0+Vsin(25)t+(1/2)(-9.81)t^2

EDIT -- Oh wait, I see the ramp 4m offset in the figure now.

 

[berkeman]

So as you said, you have two equations and two unknowns. How can you go about solving for V and t?

$$100(\frac{4}{5})=Vcos(25)t$$
$$-4-100(\frac{3}{5})=0+Vsin(25)t+(1/2)(-9.81)t^2$$

 

[Mr Beatnik]

Yeah I need to solve for V and t.

 

[berkeman]

Yeah I need to solve for V and t.

So have at it! What would be a good way to start?

 

[Mr Beatnik]

? I just ran through it again and it worked out. Unbelievable. I guess what thy say about walking away from the problem and coming back to it later really works. I am going to post my work. Thanks for your help.

 

[berkeman]

Great! Good job.

 

[Mr Beatnik]

Here is the work. It's a bit sloppy because I ran through it. Please excuse the mess.