Tiger Woods hits a golf ball off the edge of a 50 m high cliff, at an initial speed of 30 m/s and an initial angle = 40° from the horizontal. Neglect air resistance.

a) How long is the ball in flight?

b) How far from the base of the cliff does it land?

c) What is the ball's speed just before it hits the ground?

This is clearly a projectile motion problem. We won't be able to use the `Range' equation for part b), since the initial and final heights differ, so we better just start at the beginning.

Choose a coordinate system with y=vertical, x=horizontal, and the origin at the base of the cliff.

The equation for the y-component is

y = y₀ + v_0yt + ½at²

Now put in the condition that it reach the ground (y=0), call the height of the cliff y₀, set a = -g, and v_0y = v₀sin . Then the equation is

0 = y₀ + v₀sint - ½gt²

which is a quadratic equation for t. Plugging in the values v₀ = 30 m/s, g = 9.8 m/s², = 40°, and solving we get the two solutions

t = -1.79 s    t = +5.71 s

The first is unphysical (it is before the ball was hit - even Tiger Woods isn't that good!), so we choose the second solution: The ball was in flight for 5.71 s.

b) Now that we know the flight time, we can plug it into the x-equation:

x = x₀ + v_0xt

where x₀=0 here, and v_0x = v₀cos, so

x = (30 m/s)cos(40°)(5.71 s) = 131 m

c) The x-component of the velocity is a constant (no horizontal acceleration), and so

v_x = v_ox = v₀cos = (30 m/s)cos(40°) = 23.0 m/s

The y-component of the velocity is

v_y = v_0y - g t

= (30 m/s)sin(40°) - (9.8 m/s²)(5.71 s)

= -36.8 m/s

The speed is then

v = [v_x² + v_y²]^1/2

= [(23.0 m/s)² + (-36.8 m/s)²]^1/2

= 43.3 m/s

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