The maximum height reached by the squirrel : 0.479 m
Further explanationGiven
vo= 4 m/s
θ = 50 °
Required
The maximum height
Solution
Parabolic motion :
[tex]\tt h_{max}=\dfrac{v_o^2sin^2\theta}{2.g}[/tex]
Input the value
[tex]\tt h_{max}=\dfrac{4^2\times (sin~50)^2}{2\times 9.8}\\\\h_{max}=0.479~m[/tex]
or you can use
Find t from vt= vo sin θ - gt(negative sign=against gravity)⇒vt=0 at peak(the maximum height)
and input t to vertical component : y=voy.t-1/2gt²
A small sphere of mass m and charge –q is released from rest at point T. If the electric potentials at points S and T are VS and VT, respectively, what is the speed of the sphere when it reaches point S? Ignore the effects of gravity.
(A) 2q/m(Vs + VT)
(B) 4q/m(Vs + VT)
(C) q/2m(Vs - VT)
(D) q/2m (Vs + VT)
(E) 2q/m(Vs - VT)
Answer:
(E) √[2q/m(Vs - VT)]
Explanation:
Since the charge -q moves from VT to VS, the potential difference is VT - VS.
The work done in moving the charge q across a potential difference V is given by W = qV.
Now, the work done in moving the charge -q across that potential difference VT - VS is thus W = -q(VT - VS) = -q[-(VS - VT)] = q(VS - VT). This work equals the sphere's kinetic energy and kinetic energy equals K = 1/2mv² where m = mass of sphere and v = speed of sphere.
So, K = W
1/2mv² = q(VS - VT)
v² = 2q/m(VS - VT)
taking square root of both sides, we have
v = √[2q/m(Vs - VT)]