A tiger leaps horizontally from a 5.5 m high rock with a speed of 3.1 m/s. How far from the base of the rock will she land?
How long does it take to fall 5.5m?
d=1/2 g t^2 solve for time
how far can the tiger go at 3.1m/s in time t?
3.3m
To find out how far from the base of the rock the tiger will land, we can use the principles of projectile motion. The horizontal distance traveled by the tiger can be determined using the equation:
Distance = Speed × Time
In this case, we need to determine the time it takes for the tiger to reach the ground. Since the tiger is only moving horizontally, the vertical component of its motion does not affect the horizontal distance.
To find the time of flight, we can use the equation for vertical displacement:
Vertical Displacement = Initial Vertical Velocity × Time + (1/2) × Acceleration × Time^2
In this case, the initial vertical velocity is 0 m/s (since the tiger is only leaping horizontally) and the acceleration is due to gravity, which is approximately 9.8 m/s^2. The vertical displacement is the height of the rock, which is 5.5 m. Rearranging the equation, we get:
5.5 m = (1/2) × 9.8 m/s^2 × Time^2
Simplifying this equation, we find:
Time^2 = (2 × 5.5 m) / 9.8 m/s^2
Time^2 = 1.12 s
Taking the square root of both sides, we find:
Time = 1.06 s (rounded to two decimal places)
Now that we know the time, we can use the equation for horizontal distance to find the answer:
Distance = Speed × Time
Distance = 3.1 m/s × 1.06 s
Distance = 3.286 m
Therefore, the tiger will land approximately 3.29 meters from the base of the rock.