A coconut thrown horizontally at 20 m/s from the top of a hill 63 m high. How far from the base of the hill does the coconut hit the ground?
72
falls 63 m
63 = 4.9 t^2
calculate t which is time in air
now
range = 20 t
the end.
Why did the coconut go to school? To improve its "coco-nometry" skills, of course! But let's crunch the numbers and find the answer to your question.
We can solve this problem using the concept of projectile motion. Since the coconut is thrown horizontally, it does not have any initial vertical velocity. Therefore, we can ignore the vertical component of the coconut's motion and focus on its horizontal motion.
The time it takes for the coconut to hit the ground can be found using the equation:
time = distance / velocity
Since the only horizontal distance it is covering is the distance from the base of the hill, we can write the equation as:
time = distance from base / horizontal velocity
Given that the horizontal velocity is 20 m/s, we can rearrange the equation to solve for the distance from the base:
distance from base = time * horizontal velocity
To find the time it takes to hit the ground, we can use the equation for vertical motion:
y = y0 + v0 * t + 0.5 * a * t^2
In this equation, y represents the vertical position, y0 is the initial height, v0 is the initial vertical velocity, a is the acceleration due to gravity (approximately -9.8 m/s^2), and t is the time.
Plugging in the numbers:
0 = 63 m + 0 * t + 0.5 * (-9.8 m/s^2) * t^2
Simplifying the equation, we get:
0.5 * (-9.8 m/s^2) * t^2 = -63 m
Dividing both sides by 0.5 * (-9.8 m/s^2), we get:
t^2 = 12.8571 s
Taking the square root of both sides, we find:
t ≈ 3.5884 s
Plugging this value into the equation for the distance from the base, we get:
distance from base ≈ 3.5884 s * 20 m/s ≈ 71.768 m
Therefore, the coconut will hit the ground approximately 71.768 meters from the base of the hill.
To find the horizontal distance the coconut travels before hitting the ground, we can use the equation for horizontal motion:
Distance = Initial Velocity × Time
Since the coconut is thrown horizontally, the initial velocity in the horizontal direction is equal to the horizontal component of the initial velocity, which remains constant throughout the motion. We do not consider any horizontal acceleration, as there is no force acting horizontally on the coconut.
The vertical motion of the coconut can be modeled using the equations of motion:
Final Velocity (vf) = Initial Velocity (vi) + Acceleration (a) × Time
Distance (d) = Initial Velocity (vi) × Time + 0.5 × Acceleration (a) × Time²
In this case, the initial velocity in the vertical direction is 0 m/s, and the acceleration due to gravity (a) is -9.8 m/s² (assuming upward as positive and downward as negative).
To find the time it takes for the coconut to hit the ground, we can substitute the given values into the equation for vertical motion:
0 = 0 + (-9.8) × Time
Time = 0 / (-9.8)
Time = 0
The time taken for the coconut to hit the ground is 0 seconds, indicating that the coconut hits the ground instantaneously.
Now we can find the horizontal distance traveled by substituting the initial horizontal velocity (20 m/s) into the equation for horizontal motion:
Distance = Initial Velocity × Time
Distance = 20 m/s × 0 s
Distance = 0 meters
Therefore, the coconut hits the ground at the base of the hill, a distance of 0 meters from the base.
To find the distance from the base of the hill where the coconut hits the ground, we can use the kinematic equation for horizontal motion:
\(d = v \cdot t\)
where:
- \(d\) is the distance traveled
- \(v\) is the initial horizontal velocity (20 m/s)
- \(t\) is the time taken to reach the ground
First, we need to determine the time taken for the coconut to reach the ground. We can use the vertical motion equation for free fall:
\(h = \frac{1}{2} g \cdot t^2\)
where:
- \(h\) is the height of the hill (63 m)
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s²)
- \(t\) is the time taken to reach the ground
Rearranging the equation to solve for \(t\), we get:
\(t = \sqrt{\frac{2h}{g}}\)
Substituting the given values, we find:
\(t = \sqrt{\frac{2 \cdot 63}{9.8}}\)
Now, we can calculate the horizontal distance traveled by the coconut:
\(d = v \cdot t\)
\(d = 20 \cdot \sqrt{\frac{2 \cdot 63}{9.8}}\)
Evaluating this expression will give us the desired answer.