Suppose a hit golf ball went a distance of 471 m. Suppose the ball was shot horizontally off a cliff at 80.0 m/s. What is the height of the cliff?

Question

Benjamin Asiedu · Accepted Answer

#169.8"m"#

The horizontal component of velocity is constant so we can get the time of flight from:

#t=s/v=471/80=5.89"s"#

Now we can use the vertical component of motion. The initial component of the velocity is zero so we can write:

#h=1/2"g"t^2#

#:.h=1/2xx9.8xx5.89^2=169.8"m"#

Elena Albarran · Answer

undefined To find the height of the cliff, we can use the kinematic equation for projectile motion, specifically the equation for vertical motion:

$$ d = v_i t + \frac{1}{2} g t^2 $$

Where:
- $ d $ is the vertical displacement (which is the height of the cliff).
- $ v_i $ is the initial vertical velocity, which is 0 m/s since the ball is shot horizontally.
- $ t $ is the time taken for the ball to hit the ground.
- $ g $ is the acceleration due to gravity, approximately $ 9.8 \, \text{m/s}^2 $.

First, we need to find the time $ t $ it takes for the ball to hit the ground. We can use the horizontal distance traveled and the initial horizontal velocity for this:

$$ d = v_i t $$
$$ 471 \, \text{m} = (80.0 \, \text{m/s}) \times t $$
$$ t = \frac{471 \, \text{m}}{80.0 \, \text{m/s}} $$
$$ t \approx 5.8875 \, \text{s} $$

Now that we have the time $ t $, we can use it to find the height of the cliff:

$$ d = \frac{1}{2} g t^2 $$
$$ d = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (5.8875 \, \text{s})^2 $$
$$ d \approx \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times 34.615 \, \text{s}^2 $$
$$ d \approx \frac{1}{2} \times 9.8 \times 34.615 \, \text{m} $$
$$ d \approx \frac{9.8 \times 34.615}{2} \, \text{m} $$
$$ d \approx 169.931 \, \text{m} $$

Therefore, the height of the cliff is approximately $ 169.931 \, \text{m} $.