Y^2 = Yo^2 + 2g*h = 0
Yo^2 = -2g*h = -2*(-9.8)*23.2 = 454.72
Yo = 21.32 m/s = Ver. component of initial velocity.
Tan A = Yo/Vo = 21.32/57.4 = 0.37150
A = 20.38o
a. Vo = 57.4m/s[20.4o]
Xo = 57.4*Cos20.4 = 53.8 m/s = Hor.
component of initial velocity.
V = Xo + Yi = 53.8 + 0 = 53.8 m/s at highest point.
KE = 0.5*M*V^2 = 0.5*0.0505*53.8^2 =
73.1 J.
b. Y^2 = Yo^2 + 2g*h=0 + 19.6(23.2-7.97)
= 298.5
Y = 17.3 m/s
V = Xo+Yi = 53.8 + 17.3i
V = sqrt(53.8^2+17.3^2)
A 50.5-g golf ball is driven from the tee with an initial speed of 57.4 m/s and rises to a height of 23.2 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 7.97 m below its highest point?
2 answers
Correction: Replace Tan A = Yo/Vo with
sin A = Yo/Vo = 21.32/57.4 = 0.37143
A = 21.8o = Launch angle.
a. Vo = 57.4m/s[21.8o]
Xo = 57.4*Cos21.8 = 53.3 m/s = Hor.
component of initial velocity.
V = Xo+Yi = 53.3 + 0 = 53.3 m/s at highest point.
KE = 0.5*M*V^2 = 0.5*0.0505*53.3^2 =
71.7 J.
b. Y^2 = Yo^2 + 2g*h=0 + 19.6(23.2-7.97)= 298.5
Y = 17.3 m/s
V = Xo+Yi = 53.3 + 17.3i
V = sqrt(53.3^2+17.3^2) =
sin A = Yo/Vo = 21.32/57.4 = 0.37143
A = 21.8o = Launch angle.
a. Vo = 57.4m/s[21.8o]
Xo = 57.4*Cos21.8 = 53.3 m/s = Hor.
component of initial velocity.
V = Xo+Yi = 53.3 + 0 = 53.3 m/s at highest point.
KE = 0.5*M*V^2 = 0.5*0.0505*53.3^2 =
71.7 J.
b. Y^2 = Yo^2 + 2g*h=0 + 19.6(23.2-7.97)= 298.5
Y = 17.3 m/s
V = Xo+Yi = 53.3 + 17.3i
V = sqrt(53.3^2+17.3^2) =
2 answers