A projectile is fired straight upward at
167 m/s. How fast is it moving at the instant it reaches the top of its trajectory? Answer in units of m/s
How fast is it moving at the instant it reaches the top of its trajectory if the projectile is fired upward at 13◦ from the horizontal? Answer in units of m/s
To find the speed of the projectile when it reaches the top of its trajectory, we can use the concept of projectile motion.
1. In the first case where the projectile is fired straight upward, the speed at the top of its trajectory will be zero. This is because at the highest point, the vertical component of velocity becomes zero before the projectile starts to fall back down.
Therefore, the speed of the projectile when it reaches the top of its trajectory (fired straight upward) is 0 m/s.
2. In the second case where the projectile is fired at an angle of 13 degrees from the horizontal, we need to consider both the horizontal and vertical components of velocity.
Given:
Initial velocity (v0) = 167 m/s
Launch angle (θ) = 13 degrees
To find the vertical component of velocity (v_vertical) when the projectile reaches the top of its trajectory, we use the equation:
v_vertical = v0 * sin(θ)
v_vertical = 167 m/s * sin(13°) = 167 m/s * 0.224 = 37.35 m/s
Therefore, the speed of the projectile when it reaches the top of its trajectory (fired at 13 degrees) is approximately 37.35 m/s.
To answer the first question, we can use the concept of projectile motion to find the speed of the projectile at the top of its trajectory. Since the projectile is fired straight upward, its initial velocity will be vertically upwards and its acceleration will be the acceleration due to gravity acting downward. At the top of its trajectory, the projectile will momentarily come to a stop before it starts to fall back down. At this point, its velocity will be zero.
To find the speed at the top of the trajectory, we can use the equation for the final velocity of an object in free fall from rest:
v² = u² + 2as
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), and s is the displacement.
Rearranging the equation to solve for u, we get:
u = sqrt(v² - 2as)
Substituting the given values, we have:
u = sqrt(0 - 2 * 9.8 * s)
Since the displacement at the top of the trajectory is zero, we have:
u = sqrt(0 - 2 * 9.8 * 0)
This simplifies to:
u = sqrt(0)
So, the speed of the projectile at the instant it reaches the top of its trajectory is 0 m/s.
Now, let's move on to the second question. In this case, the projectile is fired upward at an angle of 13° from the horizontal. We need to find the speed of the projectile at the highest point of its trajectory.
To determine the upward component of the initial velocity, we can use trigonometry. The vertical component of the initial velocity can be found using the equation:
v_y = u * sin(θ)
where v_y is the vertical component of the velocity, u is the initial velocity, and θ is the angle from the horizontal.
Substituting the given values, we have:
v_y = 167 m/s * sin(13°)
Calculating this, we find:
v_y = 167 m/s * 0.2249
v_y ≈ 37.623 m/s
At the highest point, the vertical velocity component is 0 m/s, and the projectile is momentarily at rest before it starts to fall back down. The horizontal component of the velocity remains constant throughout the motion and does not affect the speed at the top.
Therefore, the speed of the projectile at the instant it reaches the top of its trajectory is approximately 37.623 m/s.
1. Zero.
2. Zero.
After the initial velocity, the object
slows down and continues to slow down
until it reaches its' max. ht. where the
velocity is zero. This is true at any
launching angle.