A man runs with a speed of 6 m/s off a platform that is 10m above the water. How fast is he moving when he heats the water
To find the speed at which the man hits the water, we can use the principles of kinematics and the equations of motion.
Let's assume that the initial velocity of the man is 6 m/s (as given in the question), and the acceleration due to gravity is 9.8 m/s² (assuming no air resistance and neglecting any other forces).
We can start by calculating the time it takes for the man to reach the water from the platform's height. To do that, we can use the kinematic equation:
y = v₀t + 1/2at²
Where:
y = displacement or height (which is 10m in this case)
v₀ = initial velocity (6 m/s)
t = time taken
a = acceleration due to gravity (-9.8 m/s²)
Plugging in the values, we get:
10 = 6t + 1/2 * (-9.8) * t²
Simplifying the equation:
10 = 6t - 4.9t²
Rearranging the equation to a quadratic form:
4.9t² - 6t + 10 = 0
Now, we can solve this quadratic equation using any suitable method, such as factoring, completing the square, or using the quadratic formula.
Solving the equation, we find that t ≈ 1.62 seconds (rounded to two decimal places).
Finally, to find the speed at which the man hits the water, we use the equation:
v = v₀ + at
Plugging in the values:
v = 6 + (-9.8) * 1.62
Calculating the expression:
v ≈ -3.98 m/s
The negative sign indicates that the man hits the water with a downward motion or in the opposite direction of his initial velocity. Therefore, the man is moving at approximately 3.98 m/s (rounded to two decimal places) when he hits the water.