Determine the maximum height above ground, hmax in meters, attained by the golf ball.?
A famous golfer strikes a golf ball from the ground, imparting an initial velocity described by the vector ( \mathbf{v_0} = v_{0x} \mathbf{i} + v_{0y} \mathbf{j} ). For this scenario, the initial horizontal and vertical components of the velocity are ( v_{0x} = 6 \, \text{m/s} ) and ( v_{0y} = 3.4 \, \text{m/s} ), respectively. Assume the ball moves without air resistance, and its motion is analyzed using a Cartesian coordinate system with the origin located at the ball's initial position. Determine the maximum height above the ground, ( h_{\text{max}} ), attained by the golf ball in meters.
2 Answers
Em = Ep + Ec = m.g.z + m.V²/2 = cste
g.z○ + V○x²/2 + V○z²/2 = g.z○ + g.h + V○x²/2 + 0
h = V○z²/(2.g) = 3,4^2/2/9,81 = 0,589194699 = 0,59 m
◙◙◙◙◙
Az = d²z/dt² = - g
Vz = dz/dt = V○z - g.t
Dz = z-z○ = V○z.t - g.t²/2
▬
Vz(T) = 0
T = V○z/g
Dz(T) = h = V○z²/(2.g) = 0,59 m
use vf = vi + at and remedy for vi whilst t = 4.5 and a = -32 and vf = 0 using fact its at its maximum then use that vi (that's a hundred and forty four ft/s) interior the equation d = vit + a million/2at^2 t = 5.7 (4.5+a million.2) very final answer could be 3 hundred.ninety six ft
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