Solve Kinematics Problem: 3x Max Height, Find Angle

AI Thread Summary
To determine the angle when the reach is three times the maximum height, the formulas for reach and maximum height were combined, resulting in the equation 3sin(θ)/2 = 2cos(θ). The discussion confirms that the approach to isolate θ was correct. The final step involves using a calculator to find arctan(4/3) to solve for the angle. This method effectively addresses the kinematics problem posed. The solution is confirmed as accurate and straightforward.
Johnny Blade
Messages
30
Reaction score
0
What is the angle if the reach is three (3) times the maximum height.

I put both of the reach and max height formula together to isolate theta

p=\frac{V_{0}^{2}sin(2\theta)}{-g} and h_{max}=\frac{V_{0}^{2}sin^{2}\theta}{-2g}

At the end it gave me this:

\frac{3sin\theta}{2}=2cos\theta

Am I getting rusty on my trigonometry? How would you solve this?
 
Physics news on Phys.org
Johnny Blade said:
What is the angle if the reach is three (3) times the maximum height.

I put both of the reach and max height formula together to isolate theta

p=\frac{V_{0}^{2}sin(2\theta)}{-g} and h_{max}=\frac{V_{0}^{2}sin^{2}\theta}{-2g}

At the end it gave me this:

\frac{3sin\theta}{2}=2cos\theta

Am I getting rusty on my trigonometry? How would you solve this?
your solution was right, all u have to do then is to plug it into your calculator and calculate arctan4/3
 
I see it. Thank you very much.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top