Karan Gupta
Last Activity: 7 Years ago
At the top,
V (y) = 0. Only V (x) is there. Also, we know that there is no acceleration in X direction. Hence, V(x) = U(x).
Y- direction (vertical):
Using kinematic equation, Vsq = Usq + 2gS, we put V(y)=0. Usq = 2*10*20. Usq = 400. U (y) = 20 m/s
X- direction.
Kinetic energy = 1/2mV*V = 1/2mV(x)*V(x) [Since V(y) = 0]
Plugging in the values, 200 = 1/2*1*V(x)*V(x)
V(x)*V(x) = 400.
Velocity of projection = sqrt[U(x)*U(x) + U(y)*U(y)]
= sqrt[400 + 400]
= 20√2