Final answer:
To calculate the stuntman's horizontal velocity, first find the fall time using the kinematic equation for vertical motion and then divide the horizontal distance by this time, yielding an approximate horizontal velocity of 31.35 m/s.
Step-by-step explanation:
The question involves calculating the horizontal velocity of a stuntman who projects himself from a height and lands at a certain horizontal distance. This is a classic problem in projectile motion in physics, specifically kinematics.
To calculate the horizontal velocity (Vx) with which the stuntman projected himself, we use the formula:
Vx = distance / time
First, we determine the time (t) it takes for the stuntman to fall the given vertical distance (50 meters) under the influence of gravity, ignoring air resistance. Using the kinematic equation for vertical motion:
y = (1/2)gt²
Where y is the vertical distance, g is the acceleration due to gravity (approximately 9.81 m/s²), and t is the time. Solving for t gives us:
t = √(2y/g)
Substituting the given values:
t = √(2*50/9.81) ≈ 3.19 seconds
Now that we have the time, we can find the horizontal velocity:
Vx = 100 meters / 3.19 seconds ≈ 31.35 m/s
Therefore, the horizontal velocity with which the stuntman projected himself is approximately 31.35 m/s.