A projectile is fired straight upward at
143 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 12◦
from the ho
It only has vertical speed.
That speed is zero at the top.
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Now the real one
at top vertical velocity is zero
HOWEVER the is NO horizontal force, so horizontal component is constant for the whole trip
u = 143 cos 12 = 140 m/s
Well, well, well, looks like we have a projectile on the loose! Let's see how fast it's moving at the top of its journey, shall we?
When a projectile reaches the top of its trajectory, its vertical velocity becomes zero. That's right, it takes a little breather up there! So, the speed at the instant it reaches the top is a big fat zero m/s. It's just hanging out, taking in the view, enjoying the moment. Talk about a lazy projectile!
Now, if the projectile is fired upward at an angle of 12 degrees from the horizontal, things get a little more interesting. We're not dealing with a straight-up journey anymore, folks! This time, we have both vertical and horizontal motion to consider.
To find the speed at the instant the projectile reaches the top, we need the vertical component of the velocity. The horizontal component doesn't affect the speed at the top.
So, let's convert that 143 m/s speed into horizontal and vertical components. The vertical component can be found using trigonometry. Since the angle is 12 degrees, we can go ahead and say that the vertical component is 143 * sin(12) m/s. Plug that into your calculator, and you'll have your answer!
But hey, I'm just a clown bot, not a math whiz. So, I'll leave the calculations up to you. Enjoy juggling with the numbers!
To calculate the speed of the projectile at the instant it reaches the top of its trajectory, we can use the fact that the velocity at the highest point is zero.
The initial velocity of the projectile is 143 m/s, and the acceleration due to gravity is 9.8 m/s². The projectile's velocity will decrease by 9.8 m/s every second until it reaches the top.
Using the kinematic equation:
Vf = Vi + at
where:
Vf = final velocity (0 m/s)
Vi = initial velocity (143 m/s)
a = acceleration (-9.8 m/s²)
t = time (unknown)
We can rearrange the equation to solve for time:
0 = 143 - 9.8t
Solving for t:
9.8t = 143
t = 143/9.8
t ≈ 14.59 s
Therefore, it takes approximately 14.59 seconds for the projectile to reach the top of its trajectory.
To calculate the speed of the projectile at this instant, we can use the equation:
Vf = Vi + at
where:
Vi = initial velocity (143 m/s)
a = acceleration (-9.8 m/s²)
t = time (14.59 s)
Vf = 143 - 9.8(14.59)
Vf ≈ 143 - 142.742
Vf ≈ 0.258 m/s
Therefore, the projectile is moving at a speed of approximately 0.258 m/s when it reaches the top of its trajectory.
Now, let's calculate the speed of the projectile at the instant it reaches the top of its trajectory if it is fired upward at a 12° angle from the horizontal.
To find the vertical component of the initial velocity, we can use trigonometry:
Viy = Vi * sin(θ)
where:
Viy = initial vertical component of velocity
Vi = initial velocity (143 m/s)
θ = angle (12°)
Viy = 143 * sin(12°)
Viy ≈ 143 * 0.208
Viy ≈ 29.864 m/s
Now we can use the same method as earlier to find the speed at the highest point. Since there is no horizontal acceleration, the horizontal component of the velocity remains constant throughout the flight.
Using the kinematic equation:
Vf = Vi + at
where:
Vf = final velocity (0 m/s)
Vi = initial vertical component of velocity (29.864 m/s)
a = acceleration (-9.8 m/s²)
t = time (unknown)
We can rearrange the equation to solve for time:
0 = 29.864 - 9.8t
Solving for t:
9.8t = 29.864
t = 29.864/9.8
t ≈ 3.05 s
Therefore, it takes approximately 3.05 seconds for the projectile to reach the top of its trajectory.
To calculate the speed of the projectile at this instant, we can use the equation:
Vf = Vi + at
where:
Vi = initial vertical component of velocity (29.864 m/s)
a = acceleration (-9.8 m/s²)
t = time (3.05 s)
Vf = 29.864 - 9.8(3.05)
Vf ≈ 29.864 - 29.924
Vf ≈ -0.06 m/s
Therefore, the projectile is moving at a speed of approximately -0.06 m/s (downward direction) when it reaches the top of its trajectory.
To find the speed of the projectile at the instant it reaches the top of its trajectory, we need to consider the kinematics equations for motion in one dimension.
In this case, the projectile is moving straight upward, which means we need to consider only the vertical motion. The initial velocity of the projectile is 143 m/s, and we want to determine the velocity at the top of the trajectory.
At the top of the trajectory, the velocity of the projectile momentarily becomes zero before it starts to fall back down. This means that the vertical velocity component is decreasing at that point.
To find the velocity at the top, we can use the equation for final velocity in free fall motion:
vf = vi + at
In this equation, vi is the initial velocity, a is the acceleration, t is the time taken to reach the top, and vf is the final velocity.
Since the projectile is moving straight upward, the acceleration due to gravity acts in the opposite direction of the motion. So, the acceleration is equal to -9.8 m/s².
To find the time taken to reach the top, we can use the equation:
vf = vi + at
0 = 143 - 9.8t
Rearranging the equation, we can solve for t:
t = 143 / 9.8
t ≈ 14.59 seconds
Now, we can plug the values into the equation for final velocity to find the velocity at the top:
vf = 143 - 9.8 * 14.59
vf ≈ -143 m/s
The negative sign indicates that the velocity is in the opposite direction of the initial motion. However, we are only interested in the magnitude of the velocity, so we can take the absolute value:
|vf| ≈ |-143| m/s
|vf| ≈ 143 m/s
Therefore, the speed of the projectile at the instant it reaches the top of its trajectory is approximately 143 m/s.
If the projectile is fired upward at an angle of 12° from the horizontal, we can separate the motion into horizontal and vertical components. Using the same principles outlined above, we can find the speed of the projectile at the top by considering only its vertical motion.