Zero angle launch problem: Find height given inital speed and angle

AI Thread Summary
The problem involves calculating the height of a basketball's release point when thrown horizontally at 4.40 m/s, landing at a 30.0° angle with the horizontal. The relevant equations include the tangent function relating the vertical and horizontal distances and the kinematic equations for motion under gravity. The solution approach suggests finding equations for horizontal and vertical motion, then eliminating time to express height in terms of the horizontal distance. The acceleration due to gravity is considered as -9.81 m/s². The discussion emphasizes using trigonometric relationships and kinematic equations to derive the solution.
egadda2
Messages
7
Reaction score
0

Homework Statement



A basketball is thrown horizontally with an initial speed of 4.40 m/s. A straight line is drawn from the release point to the landing point making an angle of 30.0o with the horizontal. WHAT WAS THE RELEASE POINT?

Homework Equations



tan\theta = y/x
vfx= vix= 4.40 m/s
v2fy = v2iy - 2a\Deltay

a = -g = -9.81
[c]3. The Attempt at a Solution [/b]
 
Physics news on Phys.org
Hi egadda2! :smile:

(have a theta: θ :wink:)

Find the equation for x and t, and the equation for y and t, then put y = xtanθ, and eliminate t. :wink:
 
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