what is the velocity of the jumper at the end of 1.5second? How far does he fall?
V = Vo + gt = 0 + 9.8*1.5 = 14.7 m/s.
To find the velocity of the jumper at the end of 1.5 seconds, we need to know the acceleration of the jumper. Let's assume the jumper falls freely under the influence of gravity, which means the acceleration is approximately 9.8 meters per second squared (m/s^2), downward.
To calculate the velocity, we can use the following kinematic equation:
v = u + at
Where:
v = final velocity
u = initial velocity (which we assume to be zero in this case, as the jumper starts from rest)
a = acceleration
t = time
Given that the acceleration is 9.8 m/s^2 and the time is 1.5 seconds, we can substitute these values into the equation:
v = 0 + (9.8 m/s^2) * (1.5 s)
v = 0 + 14.7 m/s
Therefore, the velocity of the jumper at the end of 1.5 seconds is 14.7 meters per second (m/s), downward.
To determine how far the jumper falls, we can use another kinematic equation:
s = ut + (1/2)at^2
Where:
s = displacement (or height)
u = initial velocity
t = time
a = acceleration
Since we want to find the distance fallen by the jumper, we can assume that the initial velocity (u) is still zero. Substituting the known values into the equation:
s = 0 * 1.5 + (1/2) * (9.8 m/s^2) * (1.5 s)^2
s = 0 + (1/2) * 9.8 m/s^2 * (2.25 s^2)
s = 0 + 11.025 m
Therefore, the jumper falls approximately 11.025 meters during the 1.5-second time interval.