A person runs off a cliff. The person runs with a speed of 10 m/s and is moving to the right as he leaves the cliff. The cliff is 13 meters above a pool of water. a) Determine the acceleration of the person after leaving the cliff (x and y components). b) Determine the time it takes the person to hit the water. c) Determine the horizontal displacement of the person from leaving the cliff to hitting the water. d) Determine the velocity (x and y components) of the person just before impact. e) Determine the velocity (magnitude and direction) of the person just before impact.
a, ay=-9.8m/s^2
b. h=1/2 9.8 t^2 solve for t
c. distance=10*t
d. vx=10, vy=-9.8t
e. v= sqrt(100+vy^2)
e) the impact angle (Θ) with respect to the pool surface
... tan(Θ) = vy / vx
To solve this problem, we need to consider the motion of the person both horizontally and vertically. Let's start by analyzing the vertical motion.
a) Determine the acceleration of the person after leaving the cliff (x and y components):
Since there are no forces acting on the person in the horizontal direction, the acceleration in the x-component is zero.
In the y-component, the only force acting on the person is gravity. The person starts with an initial vertical velocity of 0 m/s at the top of the cliff. Using the equation of motion for vertical free fall:
y = y0 + V0y * t + (1/2) * a * t^2
Since the person starts from rest and the displacement (y) is 13 meters, the equation simplifies to:
13 = 0 + (1/2) * (-9.8) * t^2
Here, -9.8 m/s^2 is the acceleration due to gravity acting downwards. Solve the equation for time (t) to find the time it takes for the person to hit the water.
b) Determine the time it takes the person to hit the water:
Rearrange the equation from part (a):
t^2 = (2 * 13) / 9.8
t^2 = 26 / 9.8
t ≈ √(26 / 9.8)
Solve for t to find the time taken. It's important to note that we are assuming no air resistance is acting on the person during their fall.
c) Determine the horizontal displacement of the person from leaving the cliff to hitting the water:
Since the horizontal velocity remains constant during the motion, we can calculate the horizontal displacement using the equation:
Displacement = Velocity * Time
The horizontal velocity is given as 10 m/s, which remains constant until impact. Multiply the horizontal velocity by the time calculated in part (b) to find the horizontal displacement.
d) Determine the velocity (x and y components) of the person just before impact:
The horizontal velocity remains constant at 10 m/s. In the vertical direction, we can use the equation of motion for free fall:
Vf = Vi + a * t
The initial vertical velocity (Vi) is 0 m/s, the acceleration (a) is -9.8 m/s^2 (downwards due to gravity), and the time (t) is the value obtained in part (b). Calculate the final vertical velocity (Vf). The horizontal component of velocity remains constant at 10 m/s throughout the motion.
e) Determine the velocity (magnitude and direction) of the person just before impact:
To find the magnitude of the velocity just before impact, use the Pythagorean theorem:
Velocity magnitude = √(Vx^2 + Vy^2)
where Vx is the horizontal velocity and Vy is the vertical velocity just before impact, obtained in part (d).
To find the direction of the velocity, use trigonometry:
Velocity direction = tan^(-1)(Vy / Vx)
This gives the angle with respect to the horizontal. The direction can be expressed as the angle above or below the horizontal.