An arrow is launched vertically upward from a crossbow at 98.2 m/s. Ignoring air friction, what is its instantaneous speed at the end of 10.0 s of flight?
What is its average speed up to that moment?
How high has it risen?
What is its instantaneous acceleration 4.20 s into the flight?
vfinal=vinitial-9.8*t
average speed= (Vi+Vf)/2
hf=Vi(10)-4.9(10^2)
To calculate the answers to these questions, we can use the basic principles of motion.
1. Instantaneous speed at the end of 10.0 s of flight:
The arrow was launched vertically upward, so we can assume it is moving against gravity. The acceleration due to gravity is approximately 9.8 m/s^2 (downward).
Using the equation of motion: v = u + at
Where:
v = final velocity (unknown)
u = initial velocity = 98.2 m/s (upward)
a = acceleration = -9.8 m/s^2 (downward)
t = time = 10.0 s
Substituting the values into the equation:
v = 98.2 m/s + (-9.8 m/s^2) * 10.0 s
v = 98.2 m/s - 98.0 m/s
v = 0.2 m/s (upward)
The instantaneous speed at the end of 10.0 s of flight is 0.2 m/s (upward).
2. Average speed up to that moment:
The average speed can be calculated using the formula: average speed = total distance / total time.
Since the arrow is moving vertically upward, the total distance covered in 10.0 s is equal to the height it has reached, considering it will fall back down.
Using the equation of motion: s = ut + (1/2) a t^2
Where:
s = distance covered (unknown)
u = initial velocity = 98.2 m/s (upward)
t = time = 10.0 s
a = acceleration = -9.8 m/s^2 (downward)
Substituting the values into the equation:
s = (98.2 m/s) * (10.0 s) + (1/2) * (-9.8 m/s^2) * (10.0 s)^2
s = 982 m - 490 m
s = 492 m
Therefore, the arrow has risen 492 meters in 10.0 seconds.
Since the arrow has gone up and will come down, the average speed will be half the total distance divided by the total time.
Average speed = (1/2) * 492 m / 10.0 s
Average speed = 24.6 m/s
So, up to that moment, the average speed of the arrow is 24.6 m/s.
3. How high has it risen:
As already calculated, the arrow has risen to a height of 492 meters.
4. Instantaneous acceleration 4.20 s into the flight:
At any point during the flight, the only force acting on the arrow is gravity, which results in a constant acceleration of -9.8 m/s^2 (downward). Therefore, the instantaneous acceleration of the arrow remains constant throughout the flight.
Hence, the instantaneous acceleration 4.20 s into the flight is -9.8 m/s^2.
To answer these questions, we can use the equations of motion for objects in free fall.
1. Instantaneous speed at the end of 10.0 s of flight:
The initial upward velocity of the arrow is 98.2 m/s, and ignoring air friction, the only force acting on the arrow is gravity. Thus, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the acceleration is equal to the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative since it acts downward). Plugging in the values, we get:
v = 98.2 m/s + (-9.8 m/s^2) * 10.0 s = 98.2 m/s - 98.0 m/s = 0.2 m/s
Therefore, the instantaneous speed at the end of 10.0 s of flight is approximately 0.2 m/s.
2. Average speed up to that moment:
Average speed is defined as the total distance traveled divided by the total time taken. Since the arrow is launched vertically, we know that it will eventually fall back down to its starting position. So, the total distance traveled up and then down is zero. Therefore, the average speed up to that moment is also zero.
3. Height reached:
To determine how high the arrow has risen, we can use the equation for displacement in free fall: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Again, we use -9.8 m/s^2 for the acceleration. Plugging in the values for t = 10.0 s, we have:
s = (98.2 m/s) * (10.0 s) + (1/2) * (-9.8 m/s^2) * (10.0 s)^2
s = 982 m + (-490 m) = 492 m
Therefore, the arrow has risen approximately 492 meters.
4. Instantaneous acceleration at 4.20 s:
Since the arrow is launched vertically upward, it will experience a constant acceleration of -9.8 m/s^2 due to gravity. Therefore, the instantaneous acceleration at any point during the flight will be the same. Thus, the instantaneous acceleration at 4.20 s is approximately -9.8 m/s^2.