A man is using a bow and arrow for the first time of his life. He is 25 meters from the target and aims directly at the center of the target. Surprisingly the arrow leaves the bow with a velocity of 54 m/s directly towards the center of the target and initially the velocity is parallel to the floor. How far below the center does the arrow hit the target?
time of flight: 25m/54m/s=25/54 s
distance fell: h=1/2 g t^2
The arrow takes
t = (25 m)/(54 m/s) = 0.46296 s
to get to the target.
Calculate how far it falls in that time using
y = (g/2)*t^2
g is the acceleration of gravity. You must know its value by now.
To determine how far below the center the arrow hits the target, we need to consider the horizontal and vertical components of the arrow's motion separately.
Let's first calculate the time it takes for the arrow to hit the target.
The horizontal component of the arrow's velocity remains constant at 54 m/s throughout its flight since there is no horizontal acceleration. We can use the formula:
distance = velocity × time
Since the distance is 25 meters and the velocity is 54 m/s, we can rearrange the formula to solve for time:
time = distance / velocity
time = 25 m / 54 m/s
time ≈ 0.463 seconds
Now, let's calculate how far the arrow falls vertically during this time.
The vertical motion of the arrow is influenced by gravity, which causes it to accelerate downward. The arrow's initial vertical velocity is 0 m/s since it is parallel to the floor.
We can use the formula for vertical displacement:
vertical displacement = (initial vertical velocity × time) + (0.5 × acceleration due to gravity × time^2)
Since the initial vertical velocity is 0 m/s and the acceleration due to gravity is approximately 9.8 m/s^2, we can substitute these values into the formula:
vertical displacement = (0 × 0.463 s) + (0.5 × 9.8 m/s^2 × (0.463 s)^2)
vertical displacement ≈ 0.108 meters
Therefore, the arrow hits the target approximately 0.108 meters below the center.