A quarter is to be tossed into a glass at the exhibition. The glass is 2.30 m from the person’s hand. If the hand is at the same level as the top of the glass and the quarter is tossed at an angle of 30.0 ̊ to the horizontal, what initial velocity is required to land the quarter in the glass? The coin is in the air for 0.510 s.

Guest Oct 15, 2014

#4**+5 **

Chris asked me to take a look at this question.

Chris's answer is correct but the original velocity components are perhaps not displayed like most people would be used to.

You can use either the horizontal or the vertical component of the velocity to work this question out.

You do not need to use both.

Let the initial velocity be v. I have also take the hand to be the origin of the axis system. That is, x=0 and y=0.

$$$The vertical component is $\quad v\;sin30 = \dfrac{v}{2}$$

$$$The horizontal component is $\quad v\;cos30=\dfrac{\sqrt3\;v}{2}$$

I'll use the horizontal velocity to get my answer.

$$\\\ddot x=0\\\\

\dot x=\frac{\sqrt3\;v}{2}\\\\

x=\frac{\sqrt3\;v}{2}\times t \\\\

When \;\;t=0.51sec\;\; x=2.3\\\\

2.3=\frac{\sqrt3\;v}{2}\times 0.51 \\\\

v=\frac{2.3\times 2}{0.51\times \sqrt3}\\\\

v=5.2074\;m/s$$

Chris did not use calculus for his answer. It was not necessary if you use the horizontal component of the velocity because this component does not change. There is not horizontal force (that is considered) so there is no horizontal acceleration and horizontal velocity does not change.

If you use the vertical velocity to do this question then the answer is slightly different v=0.51*gravity = 5m/s

The difference would be due to slight 'errors' in the original numbers or a slight difference in the interpretation of the original question. Nothing worth worrying about.

Melody
Oct 15, 2014

#1**+5 **

The coin must travel a horizontal distance of 2.30 m in .510 secs. So, D/T = R and we have

2.30m / .510s = about 4.51 m/s

(Of course, this assumes the "depth" of the glass is negligible.)

CPhill
Oct 15, 2014

#2**0 **

but the answer to the question is apparently 5.21m/s. i dont see how it was gotten

Guest Oct 15, 2014

#3**+5 **

I think I see my error....I've only given the horizontal component of the velocity.

The vertical component is given by tan(30)*4.51 = 2.603m/s

So the total velocity is given by √[(4.51)^2 + (2,603)^2] = about 5.207m/s = about 5.21m/s rounded

CPhill
Oct 15, 2014

#4**+5 **

Best Answer

Chris asked me to take a look at this question.

Chris's answer is correct but the original velocity components are perhaps not displayed like most people would be used to.

You can use either the horizontal or the vertical component of the velocity to work this question out.

You do not need to use both.

Let the initial velocity be v. I have also take the hand to be the origin of the axis system. That is, x=0 and y=0.

$$$The vertical component is $\quad v\;sin30 = \dfrac{v}{2}$$

$$$The horizontal component is $\quad v\;cos30=\dfrac{\sqrt3\;v}{2}$$

I'll use the horizontal velocity to get my answer.

$$\\\ddot x=0\\\\

\dot x=\frac{\sqrt3\;v}{2}\\\\

x=\frac{\sqrt3\;v}{2}\times t \\\\

When \;\;t=0.51sec\;\; x=2.3\\\\

2.3=\frac{\sqrt3\;v}{2}\times 0.51 \\\\

v=\frac{2.3\times 2}{0.51\times \sqrt3}\\\\

v=5.2074\;m/s$$

Chris did not use calculus for his answer. It was not necessary if you use the horizontal component of the velocity because this component does not change. There is not horizontal force (that is considered) so there is no horizontal acceleration and horizontal velocity does not change.

If you use the vertical velocity to do this question then the answer is slightly different v=0.51*gravity = 5m/s

The difference would be due to slight 'errors' in the original numbers or a slight difference in the interpretation of the original question. Nothing worth worrying about.

Melody
Oct 15, 2014