In the figure, a 0.400 kg ball is shot directly upward at initial speed 79.4 m/s
ID: 1266242 • Letter: I
Question
In the figure, a 0.400 kg ball is shot directly upward at initial speed 79.4 m/s. What is its angular momentum about P, 7.73 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground?
In the figure, a 0.400 kg ball is shot directly upward at initial speed 79.4 m/s. What is its angular momentum about P, 7.73 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground?Explanation / Answer
a) at maximum height, L = r*P*sin(90)
= r*m*v
= 0 (at maximum height v = 0)
b) maximum height reached, h = u^2/(2*g) = 79.4^2/(2*9.8) = 321.7 m
half way back to the ground, v = sqrt(2*g*h/2)
= sqrt(g*h)
= sqrt(9.8*321.7)
= 56.1 m/s
L = r*m*v
= 7.73*0.4*56.1
= 173.5 kg.m^2/s
c) at maximum height, T = r*F*sin(90)
= 7.73*0.4*9.8
= 30.3 N.m
d) at halfway back to the ground, T = r*F*sin(90)
= 7.73*0.4*9.8
= 30.3 N.m
If figure is given I can give sign(+ or -) for the answers.
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