A projectile is launched at 45 degrees above the horizontal plane, How fast is it moving at the top of its curved trajectory? 141m.s?
is 141m/s your answer or the speed of the projectile?
141m/s is the speed of the object in this question
To determine the speed of the projectile at the top of its curved trajectory, we can use the principles of projectile motion.
First, let's assume that the launch speed of the projectile is "v". Since the projectile is launched at a 45-degree angle above the horizontal plane, we can split the initial velocity into its horizontal and vertical components.
The horizontal component (v_x) remains constant throughout the entire motion and can be calculated by multiplying the launch speed by the cosine of the launch angle. In this case, the angle is 45 degrees, which means cos(45) = √2 / 2. Thus, v_x = v * (√2 / 2).
The vertical component (v_y) changes due to the influence of gravity. The velocity at the top of the trajectory will be purely horizontal, meaning the vertical component will be zero. At the highest point of its path, the projectile is momentarily at rest horizontally before it starts to descend. So we can set the vertical component of the velocity to zero (v_y = 0).
The initial vertical velocity of the projectile can be determined by multiplying the launch speed by the sine of the launch angle. Since the angle is 45 degrees, sin(45) = √2 / 2. Thus, v_y = v * (√2 / 2).
Using the equations of motion, we can find the time (t) it takes for the projectile to reach the top of its trajectory. We know that the vertical displacement is zero (y = 0), the initial vertical velocity is (v_y), and the acceleration due to gravity (a) is -9.8 m/s². The equation for vertical displacement is given by:
y = v_y * t + (0.5) * a * t²
Substituting the known values, we have:
0 = (v * (√2 / 2)) * t + (0.5) * (-9.8) * t²
We can solve this quadratic equation to find the time at the top of the trajectory.
Once we know the time at the top of the trajectory, we can calculate the speed at that point using the horizontal velocity:
v_top = v_x = v * (√2 / 2)
Without specific values for launch speed, it is not possible to determine whether the speed is exactly 141 m/s. The calculated speed will depend on the launch speed used in the calculations.