An object is thrown with velocity v from the edge of a cliff above level ground. Neglect air resistance. In order for the object to travel a maximum horizontal distance from the cliff before hitting the ground, the throw should be at an angle theta with respect tot eh horizontal of
a) greater than 60 degrees above the horizontal
b) greater than 45 degrees but less than 60 degrees above the horizontal
c)greater than zero but less than 45 degrees above the horizontal
d)zero
e)greater than zero but less than 45 degrees below the horizontal
I know it's not d) or e). It's either b) or c). but I think it's c) to get the max distance.
If it were thrown from the ground, the optimum angle would be exactly 45 degrees. When thrown from a cliff, there is an added benefit to be gained by increasing the horizontal velocity component, since you get a certain amount of "free" free fall time because of the elevation. Therefore the answer is (d).
This is a good question for using intuition. It would take quite a while and a bit of calculus to do come up with a rigorous mathematical proof.
But if the answer is d) the object wouldn't go very far. Because we are expose to find the maximum horizontal distance.
So I think the answer is c)
I apologize for the confusion caused by my initial response. You are correct in your reasoning that the object would not travel very far if the answer were d) (zero).
To determine the maximum horizontal distance, we need to consider the projectile motion of the object. When neglecting air resistance, the horizontal and vertical motions are independent of each other.
The initial velocity of the object can be split into its horizontal component (v_x) and vertical component (v_y). The horizontal component remains constant throughout the motion, while the vertical component is influenced by gravity.
To achieve maximum horizontal distance, we want the object to spend the maximum amount of time in the air. This means we want v_y to take the least amount of time to reach zero, so that gravity can then act on the object to bring it back down to the ground.
To find the angle that results in the least amount of time for v_y to reach zero, we need to analyze the vertical motion. The vertical motion can be described by the equation:
y = v_y*t - (1/2)*g*t^2
Where:
y is the vertical displacement (height),
v_y is the initial vertical velocity,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time.
In order to reach the maximum distance before hitting the ground, the total time of flight (t_total) should be as long as possible. The total time of flight can be found by setting y equal to zero:
0 = v_y*t_total - (1/2)*g*t_total^2
Rearranging this equation, we have:
t_total = (2*v_y)/g
Now, if we substitute v_y = v*sin(theta) (where v is the magnitude of the initial velocity and theta is the angle with respect to the horizontal), we get:
t_total = (2*v*sin(theta))/g
To maximize the total time of flight, we need to maximize the value of sin(theta). The maximum value of sin(theta) is 1, which occurs when theta = 90 degrees. However, in this scenario, we are throwing the object from a cliff, meaning the initial vertical velocity (v_y) is zero.
Therefore, the object would not travel any distance horizontally if thrown at an angle of 90 degrees. We need to find a smaller angle that maximizes the horizontal distance.
For any angle less than 90 degrees, sin(theta) will always be less than 1. This means that the horizontal distance will be greater than zero. Therefore, the correct answer is option b) greater than 45 degrees but less than 60 degrees above the horizontal.
To summarize, to achieve maximum horizontal distance for the object thrown from the cliff, it should be thrown at an angle between 45 and 60 degrees above the horizontal.
Yes, you are correct that the answer is b) greater than 45 degrees but less than 60 degrees above the horizontal to achieve the maximum horizontal distance.
When an object is thrown at an angle, it follows a parabolic trajectory. The range (or horizontal distance) that the object travels depends on both the initial velocity and the launch angle.
To find the launch angle that gives the maximum range, we can consider the components of the initial velocity: the vertical component (v_y) and the horizontal component (v_x).
The time taken for the object to reach the ground is the same for both cases, regardless of the launch angle, as long as the initial vertical velocity and the acceleration due to gravity remain constant. Therefore, the object will spend the same amount of time in the air in both cases.
The horizontal distance traveled is determined by the horizontal component of the initial velocity (v_x) and the time in the air. The longer the object stays in the air, the farther it will travel horizontally.
When the object is launched at 45 degrees, the initial velocity is equally divided between the horizontal and vertical components. This means that the object spends an equal amount of time going up and coming down, resulting in maximum range for a given initial velocity.
However, when the object is launched from a height (such as a cliff), the additional vertical component of the initial velocity allows it to spend more time in free fall before hitting the ground. This extra free fall time increases the horizontal distance traveled.
Therefore, to maximize the range, the launch angle should be greater than 45 degrees but less than 60 degrees above the horizontal.