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Grade: 12
        The kinetic energy at the highest point of the parabola traced by a projectile is 204 j.The maximum height reached by it in 20 m.if it`s mass is 1 kg,the velocity of projection is( g=10 m/sq metre).  Options 1)20√2 2)40 3)√20 4)10√2
2 years ago

Answers : (1)

Karan Gupta
51 Points
							
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
2 years ago
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