A diver jumps into a 10 meter deep pool from a 5 meter platform with an initial upward velocity of 3.2 meters per second. What will the divers maximum height be?
After how many second will the diver hit the water?
If the water were drained from the pool, and I launched a ball from the platform at the same initial velocity, when would the ball hit the bottom of the pool?
You must have seen the general equation for this type of question
in your case you have
height = -4.9t^2 + 3.2t + 5
d(height)/dt = -9.8t + 3.2
= 0 for a max of height
9.8t = 3.2
t = 32/98 = 16/49 seconds
at that time,
height = -4.9(16/49)^2 + 3.2(16/49) + 5
= 5.52 m
For the second part, we would have to know how deep the pool is.
When they speak of a 10m pool, like in the Olympics, they do not refer to the depth of the pool, but rather the height of the tower for diving.
It does not mean that the pool is 10 m deep!
If the question was erroneously meant as such, then solve
-4.9t^2 + 3.2t + 15 = 0
To find the diver's maximum height, you can use the kinematic equation for vertical motion. The equation is:
v^2 = u^2 + 2as
Where:
- v is the final velocity (which is 0 at the topmost point)
- u is the initial velocity
- a is the acceleration due to gravity (approximately 9.8 m/s^2)
- s is the displacement
In this case, the initial velocity is 3.2 m/s, and the displacement is the maximum height. Since we want to find the maximum height, we set the final velocity to 0.
0^2 = 3.2^2 + 2(-9.8)s
0 = 10.24 - 19.6s
Simplifying the equation:
19.6s = 10.24
s = 0.52 meters
Therefore, the diver's maximum height above the water will be 0.52 meters.
To find the time it takes for the diver to hit the water, you can use another kinematic equation:
v = u + at
Where:
- v is the final velocity (when the diver hits the water, the final velocity is -9.8 m/s)
- u is the initial velocity
- a is the acceleration due to gravity (approximately -9.8 m/s^2, negative since the motion is downward)
- t is the time
Rearranging the equation, we have:
-9.8 = 3.2 - 9.8t
Simplifying:
9.8t = 3.2 + 9.8
9.8t = 13
t = 1.33 seconds
Therefore, it will take approximately 1.33 seconds for the diver to hit the water.
If the water were drained from the pool and you launched a ball from the platform with the same initial velocity, the time it takes for the ball to hit the bottom of the pool would be the same as the time it took for the diver to hit the water. So, in this case, the ball would also take approximately 1.33 seconds to hit the bottom of the pool.