A projectile launched straight upward at 141 m/s. How fast is it moving at the top of its trajectory? Suppose it is launched upward at 45° above the horizontal plane. How fast is it moving at the top of its curved trajectory?
To calculate the speed of the projectile at the top of its trajectory, we can use the principle of conservation of energy. At the top of the trajectory, the projectile reaches its maximum height, and its kinetic energy is converted entirely into potential energy.
1. Projectile launched straight upward:
The kinetic energy (KE) of the projectile is given by:
KE = 1/2 * mv^2,
where m is the mass of the projectile and v is its velocity.
At the top of the trajectory, all of the kinetic energy is converted into potential energy (PE), so we have:
PE = mgh,
where g is the acceleration due to gravity and h is the maximum height reached by the projectile.
Since the projectile is launched straight upward, it reaches a maximum height h, after which it starts falling back down. At this point, its velocity becomes zero.
Using the principle of conservation of energy, we equate the initial kinetic energy to the potential energy at the top of the trajectory:
1/2 * mv^2 = mgh.
Canceling out the mass, we have:
1/2 * v^2 = gh.
Solving for v, we get:
v^2 = 2gh,
v = √(2gh).
Given that the projectile is launched straight upward at 141 m/s, we can substitute this value into the equation to find the speed at the top of its trajectory.
v = √(2g * h),
v = √(2 * 9.8 m/s^2 * h).
2. Projectile launched at 45° above the horizontal:
In this case, we can break the velocity into horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes due to gravity.
The initial velocity (v) of the projectile can be split into its horizontal (v_x) and vertical (v_y) components using trigonometry:
v_x = v * cosθ,
v_y = v * sinθ,
where θ is the launch angle (45°).
At the top of the curved trajectory, the vertical component of velocity becomes zero since the projectile has reached its maximum height.
v_y = 0.
The horizontal component of velocity remains unchanged throughout the motion. Thus, the speed at the top of the curved trajectory is solely determined by the horizontal component of velocity:
v = v_x.
Substituting the values into the equations for the horizontal components of velocity:
v = v_x = v * cosθ.
Given that the projectile is launched at 141 m/s and at an angle of 45° above the horizontal plane, we can substitute these values into the equation to find the speed at the top of its curved trajectory.
To find the speed of the projectile at the top of its trajectory, we can use the principles of projectile motion.
1. Projectile launched straight upward:
When the projectile is launched straight upward, its initial velocity is solely in the vertical direction and has no horizontal component. The acceleration acting on the projectile is due to gravity, pointing downward. As the projectile reaches the top of its trajectory, its vertical velocity becomes zero before it starts falling back down.
To find the speed at the top, we can use the equation for vertical velocity:
v = u + at
Where:
- v is the final velocity (which is zero at the top)
- u is the initial velocity
- a is the acceleration due to gravity (approximately -9.8 m/s²)
Given that the initial velocity is 141 m/s and the acceleration due to gravity is -9.8 m/s², we can substitute these values into the equation:
0 = 141 + (-9.8)t
Solving for t, we get:
t = 14.39 seconds
Now, we can find the speed at the top using the equation:
v = u + at
v = 141 + (-9.8)(14.39)
v = 141 - 140.76
v ≈ 0.24 m/s
Therefore, the speed of the projectile at the top of its trajectory, when launched straight upward, is approximately 0.24 m/s.
2. Projectile launched at an angle:
When the projectile is launched at an angle above the horizontal plane, the analysis becomes a bit more complicated. The initial velocity can be separated into horizontal and vertical components. At the top of the curved trajectory, the vertical velocity component will be zero, and we can find the magnitude of the total velocity using the horizontal component.
To find the speed at the top, we need to determine the initial velocity components:
- The horizontal component (u_x) is given by: u_x = u * cos(angle)
- The vertical component (u_y) is given by: u_y = u * sin(angle)
Given that the initial velocity (u) is 141 m/s and the angle is 45°, we can substitute these values into the equations:
u_x = 141 * cos(45°) = 141 * √2 / 2 ≈ 99.49 m/s
u_y = 141 * sin(45°) = 141 * √2 / 2 ≈ 99.49 m/s
At the top of the curved trajectory, the vertical velocity (v_y) will be zero, so we only need to consider the horizontal velocity (v_x) to find the speed at the top:
v_x = u_x
v = v_x = 99.49 m/s
Therefore, the speed of the projectile at the top of its curved trajectory, when launched at a 45° angle, is approximately 99.49 m/s.
1) v=0
2) v(y) = 0 =>
v=v(x)=141 m/s