Throwing Objects - Motion in Two Dimensions

[mustafamistik]

Homework Statement

It's not homework

Relevant Equations

1/2*g*t^2

I stuck on part c and d. My attemption is attached.

 

[PeroK]

Part c) says "discuss" and I don't see any discussion. I can't work out what your answer to d) is either.

 

[mustafamistik]

Part c) says "discuss" and I don't see any discussion. I can't work out what your answer to d) is either.

Is my solving way correct for part a ? I don't know how to discuss part c, what logic should I think?

 

[PeroK]

Is my solving way correct for part a ? I don't know how to discuss part c, what logic should I think?

What is your answer for part a)?

 

[mustafamistik]

What is your answer for part a)?

It's attached.

 

[PeroK]

It's attached.

45 degrees?

That said, this is a strange problem. The answer must be close to 45 degrees, but it cannot be exactly 45 degrees, unless you take $g = 10 m/s^2$.

 

[mustafamistik]

45 degrees?

That said, this is a strange problem. The answer must be close to 45 degrees, but it cannot be exactly 45 degrees, unless you take $g = 10 m/s^2$.

I rounded it. It comes from 4,9/5*tan^2... equation.

 

[PeroK]

I rounded it. It comes from 4,9/5*tan^2... equation.

Rounding or not makes a big difference to part c).

 

[mustafamistik]

Rounding or not makes a big difference to part c).

Understood. Actually i wonder if my solution way is correct. For part a.

 

[PeroK]

Understood. Actually i wonder if my solution way is correct. For part a.

It looks okay to me.

 

[mustafamistik]

It looks okay to me.

Thanks. I don't know how to discuss part c, what logic should I think?

 

[PeroK]

Thanks. I don't know how to discuss part c, what logic should I think?

It wants calculations: angle, speed gives max height.

 

[mustafamistik]

In maximum height vertical speed must equal to 0.
$V_0*sin(45)/g= t$
$(V_0*sin(45)*t)-(1/2*g*t^2)=H_{max}$
$(V_0*sin(45)*(V_0*sin(45)/g))-(1/2*g*(V_0*sin(45)/g)^2)=H_{max}$ ...
Is this way true ? for part C.

 

[PeroK]

In maximum height vertical speed must equal to 0.
$V_0*sin(45)/g= t$
$(V_0*sin(45)*t)-(1/2*g*t^2)=H_{max}$
$(V_0*sin(45)*(V_0*sin(45)/g))-(1/2*g*(V_0*sin(45)/g)^2)=H_{max}$ ...
Is this way true ? for part C.

Can you not tidy that up? You should get a simple formula for max height if $\theta = 45°$

 

[mustafamistik]

Can you not tidy that up? You should get a simple formula for max height if $\theta = 45°$

I don't know how to do that.

 

[PeroK]

I don't know how to do that.

It's basic kinematics. $v^2 - u^2 = 2as$

Or, in this case $v_y^2 - u_y^2 = 2a_ys_y$

 

[mustafamistik]

It's basic kinematics. $v^2 - u^2 = 2as$

Or, in this case $v_y^2 - u_y^2 = 2a_ys_y$

$0-(5*sqrt(5))^2=2*H_{max}$ ?
Then $62.5=H_{max}$
But is result too big ?

 

[PeroK]

$0-(5*sqrt(5))^2=2*H_{max}$ ?
Then $62.5=H_{max}$
But is result too big ?

Isn't there a little thing called gravity involved in the calculation? And isn't the angle involved too?

 

[mustafamistik]

Isn't there a little thing called gravity involved in the calculation? And isn't the angle involved too?

Sorry. You're right.
$0-(5*sqrt(5)*sin(45))^2=2*g*H_{max}$ =4.46
Yes seems to be true .

 

[PeroK]

Sorry. You're right.
$0-(5*sqrt(5)*sin(45))^2=2*g*H_{max}$ =4.46
Yes seems to be true .

I thought you were trying to find whether $H_{max} = 3m$ or not.

 

[mustafamistik]

I thought you were trying to find whether $H_{max} = 3m$ or not.

Yes, according to my calculations $H_{max}$ bigger than 3m

 

[mustafamistik]

For part D. Radial component is zero, tangential component is $g$ of acceleration. Am i wrong?

 

[PeroK]

Yes, according to my calculations $H_{max}$ bigger than 3m

Okay, but you may want to check your calculations: $4.46m$ is too high.

 

[PeroK]

For part D. Radial component is zero, tangential component is $g$ of acceleration. Am i wrong?

I assume that's right. It's an odd question, as it has little relevance to the specific problem.

 

[mustafamistik]

I assume that's right. It's an odd question, as it has little relevance to the specific problem.

I understand. Thanks for your help.