What was the magnitude and direction of the cliff diver's starting velocity when jumping off a 57m cliff and landing in the water 4.3 seconds later, 2.15m from the cliffs edge?
The magnitude of the cliff diver's starting velocity is 7.85 m/s, and the direction is at an angle of 86° with respect to the ground.
Calculating Velocity Components
To find the magnitude of the cliff diver's starting velocity, we need to calculate the horizontal and vertical components of the velocity separately.
Vertical Component:
The cliff diver jumps off a 57m cliff and lands in the water 4.3 seconds later. Using the equation d = ut + 1/2gt², where d is the displacement, u is the initial velocity, t is the time, and g is the acceleration due to gravity, we can find the vertical component.
Given:
- d = 57m
- t = 4.3s
- g = 9.81 m/s²
Plugging these values into the equation:
57 = u(4.3) + 0.5(9.81)(4.3)²
57 = 4.3u + 90.69
4.3u = -33.69
u = -7.836 m/s
The negative sign indicates that the y component of the initial velocity is in the upwards direction.
Horizontal Component:
The horizontal distance from the cliff is 2.15m. Using the equation d = ut, we can find the horizontal component.
Given:
- d = 2.15m
- t = 4.3s
Plugging these values into the equation:
2.15 = u(4.3)
u = 0.5 m/s
Calculating Net Velocity and Direction:
The net velocity is the vector sum of the horizontal and vertical components of velocity. The magnitude of the net velocity is 7.85m/s.
To find the direction, we can use the tangent function:
tan(θ) = y/x
tan(θ) = 7.836/0.5
θ ≈ 86°
Therefore, the cliff diver jumps off with a velocity of 7.85m/s at an angle of 86° with respect to the ground.