Final answer:
To find the time the hammer was in the air, we can use the equation for vertical motion. Using the known values of the initial vertical velocity and the vertical distance, we can determine that the hammer was in the air for approximately 1.26 seconds. To find the horizontal distance traveled, we use the equation for horizontal motion. By calculating the horizontal component of the initial velocity and multiplying it by the time, we find that the hammer traveled approximately 7.25 meters in the horizontal direction.
Step-by-step explanation:
To solve this problem, we can break it down into two separate components: the vertical motion and the horizontal motion of the hammer.
Vertical motion:
We know that the hammer was caught 4.0 m above the height from which it was thrown. Using the equation for vertical motion, we can find the time the hammer was in the air. The equation is:
h = (V0y * t) + (1/2 * a * t2)
Where:
h is the vertical distance
V0y is the vertical component of the initial velocity
t is the time
a is the acceleration due to gravity (-9.8 m/s2)
Since the hammer was thrown straight up, the final vertical velocity is 0. Therefore, V0y = -13 m/s. Plugging in the known values, the equation becomes:
-4.0 = (-13 * t) + (1/2 * -9.8 * t2)
Simplifying and solving for t, we find that t = 1.26 s.
Horizontal motion:
Since there is no acceleration in the horizontal direction, the horizontal velocity of the hammer remains constant throughout its motion. The horizontal distance traveled is given by the equation:
d = V0x * t
Where:
d is the horizontal distance
V0x is the horizontal component of the initial velocity
t is the time
We can find V0x by using the equation:
V0x = V0 * cos(θ)
Where:
V0 is the initial velocity
θ is the angle of projection
Plugging in the known values, we get:
V0x = 13 * cos(62°)
Simplifying, we find that V0x = 5.76 m/s.
Now, we can find the horizontal distance traveled:
d = 5.76 * 1.26
Simplifying, we find that d ≈ 7.25 m.