A rock is thrown from the top of a cliff. It lands 2.33 seconds later 17.3 meters from the base of the cliff.
How far did it drop vertically?
What is the speed of the rock when it lands?
Assuming it was thrown horizontally, the height
h = 4.9*2.33^2 = 26.60 meters
The vertical speed vy = 9.8*2.33 = 22.83 m/s
The horizontal speed vx is just 17.3/2.33 = 7.42 m/s
The final speed is of course √(vx^2+vy^2) = 24.00 m/s
To determine how far the rock dropped vertically, we can use the kinematic equation:
y = v₀t + (1/2)gt²
where:
- y is the vertical displacement (the distance the rock dropped vertically),
- v₀ is the initial vertical velocity,
- t is the time it takes for the rock to reach the ground (2.33 seconds), and
- g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the rock is thrown horizontally, the initial vertical velocity is zero, so the equation simplifies to:
y = (1/2)gt²
Plugging in the known values:
y = (1/2)(9.8 m/s²)(2.33 s)²
Calculating this expression will give you the distance the rock dropped vertically.
To find the speed of the rock when it lands, we use the horizontal distance traveled, which is 17.3 meters, and the time it takes to reach the ground, which is 2.33 seconds.
The horizontal distance traveled is given by:
x = v₀x t
where:
- x is the horizontal displacement (17.3 meters),
- v₀x is the initial horizontal velocity, and
- t is the time it takes for the rock to reach the ground (2.33 seconds).
The initial horizontal velocity can be calculated by dividing the horizontal distance by the time:
v₀x = x / t
Plugging in the known values:
v₀x = 17.3 m / 2.33 s
Calculating this expression will give you the initial horizontal velocity.
Since there is no acceleration in the horizontal direction (assuming neglecting air resistance), the horizontal velocity remains constant. Therefore, the velocity of the rock when it lands is the same as the initial horizontal velocity.