A parachustist jumps from an airplane and freely falls y=51.8 m before opening his parachute. Thereafter, he decelerates at a=1.86 m/s^2. As he reaches the ground, his speed is 3.04 m/s.
a)How long was the parachutist in the air?
b)At what height did the parachutist jump from his plane?
See
http://www.jiskha.com/display.cgi?id=1253409013
the answer is 7,040 J
To find the answers to the given questions, we can use the equations of motion. We will start by finding the time the parachutist was in the air.
a) To find the time the parachutist was in the air, we need to find the time it took for the parachutist to reach the ground after opening the parachute. We can use the equation:
s = ut + (1/2)at^2
where:
s = distance (51.8 m),
u = initial velocity (0 m/s as the parachutist was freely falling),
a = acceleration (-1.86 m/s^2 as the parachutist decelerates),
t = time.
For the freely falling portion, the equation becomes:
51.8 = 0(t) + (1/2)(-9.8)(t^2)
51.8 = -4.9(t^2)
Rearranging the equation gives:
t^2 = -(51.8 / 4.9)
t^2 ≈ -10.612
Since time cannot be a negative value, we disregard the negative sign and take the square root:
t ≈ √10.612
t ≈ 3.26 s
Therefore, the parachutist spent approximately 3.26 seconds in the air.
b) To find the height from which the parachutist jumped, we can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity (3.04 m/s),
u = initial velocity (0 m/s),
a = acceleration (-1.86 m/s^2),
s = distance.
For the decelerating portion, the equation becomes:
(3.04)^2 = 0^2 + 2(-1.86)s
9.2416 = -3.72s
Rearranging the equation gives:
s ≈ 9.2416 / -3.72
s ≈ -2.48 m
Since height cannot be a negative value, we disregard the negative sign:
s ≈ 2.48 m
Therefore, the parachutist jumped from a height of approximately 2.48 meters.