A tiger leaps horizontally from a 15.0 m high rock with a speed of 7.10 m/s. How far (in meters) from the base of the rock will it land?
h=1/2 g t^2 solve for time t, time in the air.
Now, distance=7.1*t
3m
To find the distance from the base of the rock where the tiger will land, we need to use the equations of motion. We can use the equation relating the horizontal distance traveled, initial horizontal velocity, and time:
𝑑 = 𝑣𝑥 × 𝑡
Where:
𝑑 is the horizontal distance traveled,
𝑣𝑥 is the horizontal component of velocity, and
𝑡 is the time of flight.
Now, we need to determine the time of flight. Since the tiger leaps horizontally, the only influence on the time will be the acceleration due to gravity acting vertically downwards, which will affect the velocity in the vertical direction.
We can use the equation for vertical motion to find the time taken to reach the ground from a certain height:
𝑦 = 𝑣𝑖𝑛𝑖𝑡 + 0.5𝑎𝑡^2
Where:
𝑦 is the displacement in the vertical direction (change in height),
𝑣𝑖𝑛𝑖𝑡 is the initial vertical velocity multiplied by time (it is 0 here because the tiger leaps horizontally),
𝑎 is the acceleration due to gravity (-9.8 m/s^2), and
𝑡 is the time of flight.
Rearranging the equation, we get:
0 = 0 + 0.5 × (-9.8) × 𝑡^2
Simplifying further:
-4.9 × 𝑡^2 = 0
Since the time cannot be negative, we can disregard the negative solution and conclude that 𝑡 = 0.
This means that the tiger will take no time to land horizontally. Therefore, the horizontal distance traveled will also be zero (𝑑 = 0).
Hence, the tiger will land directly at the base of the rock.