You have a friend who is an archer and she wants to know how fast the arrows leave her bow. You have her launch an arrow horizontally, 1.5 m above the ground, and find that the arrow lands 33 m away. What is the initial velocity of the arrow as it leaves the bow?
horizontal: 33=V*time
vertical: 0=1.5-1/2 9.8 t^2
solve the first equation for time
time=33/V
Put that in the second equation for t.
Solve for V.
Dvide 33 m by the time it takes the arrow to fall 1.5 meters. Call bhat time T.
That time of flight T is obtainable from the relation:
1.5 = (g/2)T^2.
T = sqrt(3.0/g) = 0.553 s
33 m/0.553s = ___ m/s
To determine the initial velocity of the arrow as it leaves the bow, we can use the equation from basic physics:
d = v * t
Where:
- d is the distance traveled by the arrow (33 m in this case)
- v is the initial velocity of the arrow
- t is the time it takes for the arrow to travel the distance
Now, since the arrow is launched horizontally, there is no vertical acceleration. Therefore, the time it takes for the arrow to reach the ground is the same as the time it would take to fall freely from a height of 1.5 m. We can use the equation for free fall to find this time:
h = (1/2) * g * t^2
Where:
- h is the height (1.5 m in this case)
- g is the acceleration due to gravity (-9.8 m/s^2)
Rearranging the equation and solving for t:
t^2 = (2 * h) / g
t^2 = (2 * 1.5) / 9.8
t^2 = 0.306
t ≈ √0.306
t ≈ 0.553 s
Now that we have the time, we can substitute it back into the first equation to find the initial velocity of the arrow:
33 m = v * 0.553 s
Rearranging the equation and solving for v:
v = 33 m / 0.553 s
v ≈ 59.6 m/s
Therefore, the initial velocity of the arrow as it leaves the bow is approximately 59.6 m/s.