A diver of mass 143 kg takes off from the 22 m platform from a standing position by jumping upwards with a (near) vertical speed of 4.4 m/s. Calculate
i) the time taken to reach the water
ii) the initial gravitational potential energy
iii) the initial kinetic energy
Answers
The time to reach the water is 1.65 seconds.
There is more than one way to solve this type of problem. For this example, let's consider the upward travel separately from the downward travel. Since the initial velocity is 2.0 m/s upward, the time traveling upward is
2.0
m
s
9.8
m
s
2
=
0.20
s
, where 9.8 meters per second squared is the acceleration of gravity, g.
Then, we can calculate the distance the diver moves upward:
d
=
1
2
g
t
2
, where g is the acceleration due to gravity and t is the time.
d
=
4.9
m
s
2
(
0.20
s
)
2
=
0.20
m
This means that the diver falls from a height 10. 2 m not the original 10 m. Now we can use the free fall equation to calculate how long she is falling down.
10.2
=
4.9
t
2
t
2
=
2.08
s
2
t
=
1.45
s
The total time to reach the water is the 0.20 s upward plus the 1.45 s downward for a total of 1.65 s before reaching the water..
Part b.
The potential energy is P.E. = mgh
P
.
E
.
=
65
k
g
×
9.8
m
s
2
×
10
m
=
6
,
370
k
g
m
2
s
2
=
6
,
370
J
.
A joule is the SI unit equivalent to one
k
g
m
2
s
2
Part c.
The initial kinetic energy is zero. The kinetic energy is
K
.
E
.
=
1
2
m
v
2
. Since the initial velocity is zero, the kinetic energy is zero.