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?
Select one:
a. 42 m
b. 13 m
c. 66 m
d. 72 m
time to fall:
-63=0-4.9t^2
t=sqrt( )
Now, distance=20*t
To find the horizontal distance from the base of the hill where the coconut hits the ground, we need to calculate the time it takes for the coconut to fall.
The vertical distance the coconut falls is the height of the hill, which is 63 m. The initial vertical velocity is 0 m/s (since it is thrown horizontally), and the acceleration due to gravity is approximately 9.8 m/s^2.
Using the equation for vertical displacement:
h = (1/2)gt^2,
where h is the vertical displacement, g is the acceleration due to gravity, and t is the time taken.
Plugging in the given values:
63 = (1/2)(9.8)(t^2).
Rearranging the equation and solving for t^2:
t^2 = (2 * 63) / 9.8,
t^2 = 12.857,
t ≈ √(12.857), t ≈ 3.588.
The time taken for the coconut to fall is approximately 3.588 seconds.
To find the horizontal distance, we can use the equation:
distance = horizontal velocity * time.
The horizontal velocity of the coconut is given as 20 m/s, and the time we just calculated is 3.588 seconds.
Plugging the values into the equation:
distance = 20 * 3.588,
distance ≈ 71.76.
Therefore, the coconut hits the ground approximately 71.76 m from the base of the hill, which can be rounded to 72 m.
Therefore, the answer is option d. 72 m.
To find the distance from the base of the hill where the coconut hits the ground, we can use the equation of motion in the horizontal direction:
distance = velocity × time
In this case, the initial horizontal velocity of the coconut is 20 m/s, and it will continue to travel horizontally until it hits the ground. The time it takes for the coconut to hit the ground can be found using the equation of motion in the vertical direction:
h = (1/2) × g × t^2
where:
h = height (63 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation to solve for time, we have:
t^2 = (2h) / g
t^2 = (2 × 63) / 9.8
t^2 ≈ 12.853
Taking the square root of both sides to find the time, we have:
t ≈ √12.853
t ≈ 3.59 s
Now, we can substitute this time into the equation for distance:
distance = velocity × time
distance = 20 m/s × 3.59 s
distance ≈ 71.8 m
Therefore, the coconut hits the ground approximately 71.8 meters from the base of the hill. None of the provided answer options are exact, but the closest option is c. 66 m.