Calculating Time of Flight for Ball Thrown at 45° with Initial Speed 31 m/s

[unfortunate]

A ball is thrown with an initial speed of 31 m/s at an angle of 45°.The ball is thrown from a height of 10 m and lands on the ground.

(a) Find the time of flight.

http://img153.echo.cx/img153/5338/showme2mp.gif

I used the following formula:

Yf = Yo + VoT - (1/2)GT^2

0 = 0 + (31 m/s sin 45)T - (1/2)(9.8 m/s^2)T^2
0 = 21.92T - 4.9T^2
-21.92T = -4.9T^2
T = 4.47s

I'm checking my answer for the problem on my school website and it's coming back as wrong. I pretty sure I'm using the correct formula for problem. The math looks right.

 

[The Bob]

You might want to find the components of the flight in terms of horizontal and vertical flight.

The Bob (2004 ©)

 

[dextercioby]

Of course,that y_{0}\neq 0...

Daniel.

 

[stunner5000pt]

just xaplin why Yf - Y0 is zero?? DId the ball not travel some vertical distance?? I.e. Is the final height same and the initial height??

 

[dextercioby]

No,the initial height is 10 m...And he should be careful with those velocity components as well.

Daniel.

 

[unfortunate]

Well I'm confused with Y (initial) and Y (final). These are the hints that came with the problem:

We know y initial and y final are both zero since both are on the ground

 

[dextercioby]

Well I'm confused with Y (initial) and Y (final). These are the hints that came with the problem:

We know y initial and y final are both zero since both are on the ground

Have you been reading other problem?

"(...)The ball is thrown from a height of 10 m and lands on the ground.(...)".

Does this ring a big bell ?

Daniel.

 

[unfortunate]

Sorry dextercioby..

Ok so Y initial = 10m and Y final = 0m

Yf = Yo + VoT - (1/2)GT^2
0m = 10m + (31 m/s sin 45)T - (1/2)(9.8 m/s^2)T^2
T = 6.5s

Tycho is not taking my answer.

 

[The Bob]

I got 4.9 seconds for it to hit the ground.

The Bob (2004 ©)

 

[The Bob]

Unfortunate, start by finding the component of the vertical. This is all you really need to solve the question.

The Bob (2004 ©)

 

[The Bob]

Have I gone mad? Did unfortunate just post or am I dreaming?

The Bob (2004 ©)

 

[unfortunate]

Have I gone mad? Did unfortunate just post or am I dreaming?

The Bob (2004 ©)

Our posts crossed. I asked what steps you took to get 4.9s. I deleted it thinking that you were going to explain the steps.

 

[The Bob]

Our posts crossed. I asked what steps you took to get 4.9s. I deleted it thinking that you were going to explain the steps.

Fair enough well the explanation will be in the next post.

The Bob (2004 ©)

 

[The Bob]

The vertical component of the ball can be expressed in the form:

s = ut + \frac{1}{2}at^2

u = sin 45° x 31 = 22 ms^-1, a = -9.8 ms^-2 which will leave an expression for t relative to the ball's position (s).

This gives s = 22t - 4.9t² + 10 (the 10 has come from the fact that the ball has started 10 metres above the ground).

Now when s = 0 the ball has landed (yes?). This means we have a quadratic equation : 0 = 22t - 4.9t² + 10. Rearrange this so that the squared function is positive and you have 4.9t² - 22t - 10 = 0.

Then it is a case of using the quadratic equation:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This (for t instead of x) will leave you with either 4.9s or -0.41s. The negative answer cannot be what you are looking for (as this answer will give you how long ago the ball would have needed to be throw from the ground to get to 10 metres above the ground).

Hence, the answer is 4.9 seconds.

Hope that helps.

The Bob (2004 ©)

P.S. You had the idea but didn't seem to know where to take it.

 

[unfortunate]

Genius.. I'm soaking it all in. :) Now to attack the rest of the ball throw problem.

 

[The Bob]

Genius.. I'm soaking it all in. :) Now to attack the rest of the ball throw problem.

Good Luck then

The Bob (2004 ©)

 

[unfortunate]

Ok I'm having a problem finding the speed at impact. I know it's the magnitude of Vx and Vy.

I have Vx as Vi cos 45 = 21.92

I'm having trouble finding Vy.