vertical velocity of 20m/s the rock was launched with a horizontal rightward velocity component of 1.2 m/s with the height of the cliff being 25 m, how far from the base of the cliff would the rock land?
To find how far from the base of the cliff the rock would land, we need to calculate the horizontal distance travelled by the rock.
Since the rock was launched with a horizontal rightward velocity component of 1.2 m/s, the horizontal velocity remains constant throughout its flight.
Let's assume the time of flight of the rock is 't' seconds.
The horizontal distance travelled by the rock, d, can be calculated using the formula:
d = horizontal velocity × time
Given that the horizontal velocity is 1.2 m/s, and we need to find 't', we will use another formula to calculate 't'.
Let's consider the vertical motion of the rock. The rock was launched with a vertical velocity of 20 m/s, and it goes up to a height of 25 m. At the maximum height, the vertical velocity is 0 m/s.
We can calculate the time it takes for the rock to reach the maximum height using the formula:
final vertical velocity (v) = initial vertical velocity (u) + acceleration (a) × time (t)
Since the acceleration due to gravity is acting in the opposite direction to the initial velocity, the acceleration is negative (-9.8 m/s²). The vertical velocity at the maximum height is 0 m/s, and the initial vertical velocity is 20 m/s.
0 = 20 – 9.8t
Solving this equation, we find that t = 2.04 seconds (rounded to two decimal places).
Now that we know the time of flight is 2.04 seconds, we can calculate the horizontal distance travelled by the rock:
d = 1.2 × 2.04 = 2.448 m
Therefore, the rock would land approximately 2.448 meters from the base of the cliff.
Note: We have ignored air resistance in this calculation. In reality, air resistance affects the horizontal distance covered, but we have assumed it to be negligible for simplicity.