# Throwing Objects - Motion in Two Dimensions

• mustafamistik
In summary, the problem asks for an equation for the maximum height of a Vertical Slider moving at a speed of 10 m/s.

#### 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.

#### Attachments

• S3.PNG
36 KB · Views: 91
• Attemption.jpeg
37.8 KB · Views: 91
Part c) says "discuss" and I don't see any discussion. I can't work out what your answer to d) is either.

PeroK said:
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?

Delta2
mustafamistik said:
Is my solving way correct for part a ? I don't know how to discuss part c, what logic should I think?

PeroK said:
It's attached.

PeroK
mustafamistik said:
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##.

PeroK said:
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.

mustafamistik said:
I rounded it. It comes from 4,9/5*tan^2... equation.
Rounding or not makes a big difference to part c).

PeroK said:
Rounding or not makes a big difference to part c).
Understood. Actually i wonder if my solution way is correct. For part a.

mustafamistik said:
Understood. Actually i wonder if my solution way is correct. For part a.
It looks okay to me.

PeroK said:
It looks okay to me.
Thanks. I don't know how to discuss part c, what logic should I think?

mustafamistik said:
Thanks. I don't know how to discuss part c, what logic should I think?
It wants calculations: angle, speed gives max height.

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.

mustafamistik said:
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°##

PeroK said:
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.

mustafamistik said:
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##

PeroK said:
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 ?

mustafamistik said:
##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?

PeroK said:
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 .

mustafamistik said:
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.

PeroK said:
I thought you were trying to find whether ##H_{max} = 3m## or not.
Yes, according to my calculations ##H_{max}## bigger than 3m

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

mustafamistik said:
Yes, according to my calculations ##H_{max}## bigger than 3m
Okay, but you may want to check your calculations: ##4.46m## is too high.

mustafamistik said:
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.

PeroK said:
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.