a rock is dropped from a cliff and reaches the ground in 2.70s . At the same time another rock is thrown horizontally from the same height with initial x velocity of 0.95 m/s. what distance will in x will the second recover?
distance = velocity * time
To find the distance covered by the second rock in the horizontal direction (x direction), we need to first calculate the time it takes for the rock to reach the ground when only considering the vertical motion.
The first rock's vertical motion can be represented by the equation:
h = (1/2) * g * t^2
Where:
h = height (the same as the height of the cliff)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken (2.70 s)
Rearranging the equation to solve for h:
h = (1/2) * g * t^2
Substituting the values:
h = (1/2) * 9.8 * (2.70)^2
h = 35.673 m
Now, we know that the time taken for the vertical motion is 2.70 seconds, and the initial horizontal velocity (Vx) of the second rock is 0.95 m/s.
The distance covered in the horizontal direction (x direction) can be calculated using the formula:
distance = Vx * t
Substituting the values:
distance = 0.95 * 2.70
distance = 2.565 m
Therefore, the second rock will cover a distance of 2.565 meters in the horizontal direction (x direction).