66.7 m/s2
66.7 m/s2
Acceleration due to gravity is 9.8m/s/s, which is the same as 9.8m/s2. An acceleration of 9.8m/s/s means that with each passing second, the velocity of the skydiver increases by 9.8m/s. Therefore, after two seconds. a skydiver's velocity would be 19.6m/s. The acceleration will continue at 9.8m/s/s until the skydiver reaches terminal velocity, at which point the weight of the skydiver and the air resistance will be balanced, so the net force acting on the skydiver will be zero, at which point there will be no further acceleration.
Because you reach maximum velocity.
101 is.
Assuming we neglect air resistance and start from rest in the vertical plane - use v=u+at so v = 0+(10x2) assuming g = 10N/kg =20m/s In practice it will be less because air resistance will tend to slow the skydiver down.
the magnitude of the skydivers acceleration is zero as he is decelerating by opening his parachute!
66.7 m/s2
66.7 m/s2
80 m/s 2 up
Maximum speed is about 220 to 230mph and can be achieved after about 20 seconds of freefall. Normal parachute opening speed should be not greater that 120mph to avoid damage to the parachute
(4 m/s - 54 m/s)/0.75 s = -50/0.75 m/s² = -200/3 ≈ - 66.67 m/s² (negative because he is decelerating)
If the 0.75 refers to seconds, then his acceleration is -66.66... (repeating) metres per second^2.
zero
80 m/s2 up
To calculate this, you divide the change in velocity, by the time.
If d = 16*t^2 then there is no significant air resistance.
The general formula for acceleration is [(final velocity) - (initial velocity)]/(time required for the change). In this instance, (5 - 65)/0.75 = -80 meters per second per second.