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? The solution is 1.2 * 5=6 m. Where did 5 come from?
The value of 5 comes from the time it takes for the rock to fall from the cliff to the ground.
To find the time of flight, we can use the formula:
s = ut + (1/2)at^2
Where:
s = displacement (in this case, the height of the cliff, which is 25 m)
u = initial velocity (in this case, the vertical velocity of 20 m/s since the rock was launched vertically)
t = time of flight (which we need to find)
a = acceleration due to gravity (-9.8 m/s^2, assuming downward is negative)
Rearranging the formula, we have:
t = (2s / a)^(1/2)
Substituting the given values:
t = (2 * 25 / -9.8)^(1/2)
t ≈ 2.83 s (approximated to two decimal places)
Since the horizontal rightward velocity component has no effect on the vertical motion (they are independent of each other), we only need to focus on the time the rock takes to fall vertically.
Now, we can calculate the horizontal distance the rock traveled using the formula:
horizontal distance = horizontal velocity * time
Substituting the given values again:
horizontal distance = 1.2 m/s * 2.83 s
horizontal distance ≈ 3.396 m (approximated to three decimal places)
So, the rock would land about 3.396 meters from the base of the cliff. It seems there might be a rounding error in the solution provided.