Question

Tiger woods tees off fro a hill hitting a golf ball 58.4 m/s at an angle of 20.5 degrees above the horizontal. The ball lands 10.5 m below. How long is the ball airborne? What distance did the ball travel

Answer 1

By applying 2nd kinematics law in vertical direction,

Sy = Uy*t + 0.5*ay*t^2

here,

Sy = -10.5 m

Uy = U*sin = 58.4*sin(20.5 deg)

Uy = 20.45 m/s

ay = -g = -9.8 m/s^2

So, -10.5 = 20.45*t + 0.5*(-9.8)*t^2

9.8*0.5*t^2 -20.45*t - 10.5 = 0

By solving this quadratic equation,

t = time = 4.6 sec.

Now, using second kinematics law in horizontal direction,

Sx = Ux*t + 0.5*ax*t^2

here, Sx = horizontal distance = ??

Ux = U*cos = 58.4*cos(20.5 deg)

Ux = 54.7 m/s

t = 4.6 sec.

ax = horizontal acceleration = 0

So, Sx = 54.7*4.6 + 0

Sx = distance = 251.6 m

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