Determining horizontal distance travelled

Question

On Jupiter’s largest moon, gravity is 1.4 m/s2. An astronaut, standing on the edge of a 100m high cliff, hits a golf ball, sending it flying at 75 m/s [30° above the horizontal].

 

Determine the horizontal distance traveled by the golf ball

Answer 1

Looking at the vertical component of the total trajectory, we can use the kinematic equation

y = y₀ + v_0yt + 1/2*a_yt²

where we establish that the final vertical position (y = 0) is the surface of the moon and the initial vertical position (y₀ = 100 m) is the top of the cliff. The initial vertical velocity is simply v_0y = v₀sin(30) = v₀/2 = 37.5 m/s. The vertical acceleration is simply due to gravity, so a_y = -1.4 m/s².

Using the kinematic equation above and solving for t (using the quadratic formula), we can see that the ball stays in the air for approximately 2.55 s.

We can now focus on the horizontal component of the trajectory and use the kinematic equation

x = x₀ + v_0xt + 1/2*a_xt²

where x is the final horizontal position relative to x₀ = 0 m. This will be the total horizontal distance traveled and thus the solution. v_0x = v₀cos(30) = 37.5√3 m/s and there is no horizontal acceleration so a_x = 0 m/s². The important thing to note is that the time it spends in it's horizontal motion is the same amount of time that it spends in it's vertical motion.

Substitution of the required values should yield a distance of approximately 165 m.