The projectile (ball) reaches an instantaneous vertical speed (Vy) of zero at maximum height.
so, V(max height) = ¬Г(Vx)^2+(Vy)^2
in this case V(max height) = Vx, where Vy=0
The maximum height, Yf, can be solved using Vfy^2=Viy^2 + 2gy. At maximum height Vfy=0.
Answer:
power=work done÷time taken
2×5=10
10÷10=1
ans 1J per second
Answer:
Let M1 = 8 kg and M2 = 34 kg
F = M a = (M1 + M2) a
F = M2 g the net force accelerating the system
M2 g = (M1 + M2) a
a = M2 / (M1 + M2) g = 34 / (42) g = .81 g = 7.9 m/s^2