Final answer:
Using the kinematic equation for vertical displacement, assuming no initial vertical velocity, the height of the cliff from which a motorcycle stunt driver jumps if they land after 0.75 seconds is approximately 2.74 meters.
Step-by-step explanation:
To calculate the height of the cliff the motorcycle stunt driver jumps off, we can use the kinematic equation that relates velocity, acceleration, time, and displacement in the vertical direction:
s = ut + ½ at²
where:
- s is the vertical displacement (height of the cliff)
- u is the initial vertical velocity (since the driver zones off horizontally, this is 0 m/s)
- a is the acceleration due to gravity (approximately 9.81 m/s² downward)
- t is the time in the air (0.75 seconds)
Substituting the known values into the equation gives us:
s = (0 m/s)(0.75 s) + ½ (9.81 m/s²)(0.75 s)²
Clean up the formula and solve for s:
s = 0 + ½ (9.81)(0.5625)
s = ½ (5.4881)
s = 2.74405 m
So, the height of the cliff is approximately 2.74 meters.