vertical velocity of 20m/s the rock was launched with a horizontal rightward velocity component or 1.2 m/s, how far from the base of the cliff would the rock land?
To find the distance from the base of the cliff that the rock will land, we need to calculate the horizontal distance traveled by the rock.
The horizontal velocity component remains constant throughout the motion. Therefore, the horizontal distance (d) traveled by the rock can be calculated using the formula:
d = v * t,
where:
- d is the horizontal distance,
- v is the horizontal velocity component (1.2 m/s),
- t is the time of flight.
To find the time of flight, we need to use the vertical velocity component and the acceleration due to gravity. The vertical distance traveled by the rock is determined by the following equations:
d = v₀*t + (1/2)*g*t²,
where:
- v₀ is the initial vertical velocity (20 m/s),
- g is the acceleration due to gravity (-9.8 m/s²).
Since the rock is launched vertically and lands at the same level, the vertical distance traveled is zero. Therefore, we can set the equation above to zero:
0 = v₀*t + (1/2)*g*t².
Rearranging the equation gives:
(1/2)*g*t² = -v₀*t.
Simplifying further, we get:
(1/2)*g*t = -v₀.
Solving for t, we have:
t = -2*v₀/g.
Substituting the given values:
t = -2 * 20 / -9.8 = 4.08 s.
Now we can find the horizontal distance by substituting the values:
d = v * t = 1.2 m/s * 4.08 s = 4.8976 m.
Therefore, the rock will land approximately 4.8976 meters from the base of the cliff.