A projectile fired vertically up from a cannon rises 200 m before returning to the ground. If the same cannon were to fire the same projectile at an angle 45 degrees, then the maximum range would be m.
When fired straight up, the height attained is
H = V^2/(2g)
When fired at at angle A, the range is
R = (V^2/g)sin(2A)
When the angle is 45 degrees, R = V^2/g
and R = 2 H. (The H refers to the straight-up value)
Therefore the answer is 400 m.
A car is traveling along a straight road increases its speed from 30.0 m/s to 50.0 m/s in a distance of 180 m. If the acceleration is constant, how much time elapses while the car moves this distance?
To find the maximum range of the projectile fired at an angle of 45 degrees, we can use the formula for the range of a projectile:
Range = (2 * velocity^2 * sin(angle) * cos(angle)) / acceleration due to gravity
However, we don't have the initial velocity or the acceleration due to gravity. So, let's make some assumptions:
1. The initial velocity of both projectiles fired vertically and at 45 degrees is the same.
2. The acceleration due to gravity is 9.8 m/s^2.
With these assumptions, we can calculate the maximum range.
Given:
Initial vertical displacement = 200 m
Angle = 45 degrees
Acceleration due to gravity = 9.8 m/s^2
Step 1: Calculate the initial velocity of the projectile fired vertically.
Using the equation: v = sqrt(2 * g * h), where h is the vertical displacement.
v = sqrt(2 * 9.8 m/s^2 * 200 m)
v ≈ 44.29 m/s
Step 2: Calculate the maximum range of the projectile fired at an angle of 45 degrees.
Using the formula: Range = (2 * v^2 * sin(angle) * cos(angle)) / acceleration due to gravity
Range = (2 * (44.29 m/s)^2 * sin(45 degrees) * cos(45 degrees)) / 9.8 m/s^2
Range ≈ 404.09 m
Therefore, the maximum range of the projectile fired at an angle of 45 degrees would be approximately 404.09 m.
To determine the maximum range of a projectile fired at an angle of 45 degrees, we need to understand the principles of projectile motion.
When a projectile is launched at an angle, it follows a curved path known as a parabola. The range of the projectile is the horizontal distance it travels before hitting the ground again.
The maximum range occurs when the projectile is launched at an angle of 45 degrees. At this angle, the horizontal and vertical components of the velocity are equal, resulting in the maximum distance covered.
To calculate the range, we need to know the initial velocity of the projectile. However, we are given that the same projectile was fired from the same cannon, so we can assume that the initial velocity remains constant.
Let's assume the initial velocity of the projectile is V.
To find the horizontal component of the velocity (Vx) at an angle of 45 degrees, we need to use trigonometry. The horizontal velocity is given by:
Vx = V * cos(45)
Similarly, the vertical component of the velocity (Vy) can be found using:
Vy = V * sin(45)
Since the time of flight for the projectile is the same for both the vertical and the angled launch, the total time it takes to reach the ground can be found using the formula:
t = 2 * (Vy / g)
where g is the acceleration due to gravity.
Now, we can find the horizontal distance covered (range) using:
Range = Vx * t
Substituting the above expressions, we get:
Range = (V * cos(45)) * 2 * (V * sin(45)) / g
Simplifying further:
Range = (V^2 * sin(2 * 45)) / g
Since we are not given the initial velocity V, we cannot directly calculate the range. We need more information to determine the actual value of the maximum range in meters.