Help please so confused !!A swimmer bounces straight up from a diving board and falls feet first into a pool.?

Question

She starts with a velocity of 4.00 m/s, and her takeoff point is 1.10 m above the pool.
(a) How long are her feet in the air?

(b) What is her highest point above the board?

(c) What is her velocity when her feet hit the water?

Answer 1

►Taking upwards ↑ as +ve direction (my preference)



a) initial vel, vi = 4.0 m/s↑

final displacement, d = -1.10m ↓

accel (due to gravity), a = -9.80 m/s²↓



d = vi.t + ½at²

-1.10 = 4t - 4.9t² .. .. 4.9t² - 4t - 1.1 = 0 .. .. a quadratic equation to solve

►t = 1.03 s



b) max height (h) when final vel, vf = 0, a = -9.80 m/s²



vf² = vi² + 2ah

0 = 4² - 19.6h .. .. h = 16 / 19.6 .. .. ►h = 0.82 m



c) vi = -4, d = -1.10, a = -9.80, vf = ?



vf² = vi² + 2ad

vf² = 4² + (2 x -9.80 x -1.10) = 37.56 .. vf = √37.56 .. .. ►vf = 6.13 m/s



If you haven't come across quadratic equations for part (a) ..

vi = 4.0m/s↑, vf = - 6.13 m/s↓, a = -9.80↓



vf = vi + at

-6.13 = 4 - 9.8t .. .. 9.8t = 10.13 .. t = 10.13 / 9.8 .. .. t = 1.03 s

Answer 2

Given:

Initial velocity, v0 = 4.00 m/s

Height, h = 1.10 m



Solution:



a) Total time to reach the pool:



According to the equation of kinematics, the total time of flight can be determined as follows.



The vertical displacement of the diver is given by



y = y0 + v0t - (1/2)gt^2

Final displacement is zero and the initial displacement is the height of the diving board from pool. So,



0 m = 1.10 m + 4.0 m/s*t - (1/2)*9.8 m/s^2*t^2



Solve for t, we get

t = 1.03 s



The diver's feet will be in the air for a period of 1.03 seconds.



b) Highest point above the board:



According to the equation of kinematics, the highest point above the board can be determined as follows.



Time to reach the highest point can be obtained as follows.

v = v0 - gt

At the highest point, the velocity of the diver is 0 m/s. So,



0 m/s = 4.0 m/s - 9.8 m/s^2*t

solve for t, we get



t = 0.41 sec



The diver reaches the highest point above the board is



y = y0 + v0t - (1/2)gt^2

At the board, vertical displacement is zero. So,

y = 0 m + 4.00 m/s *0.41 s - (1/2)*9.8 m/s^2*0.41^2 s^2

y = 0.82 m



The diver's highest point above the board is 0.82 m.



c) Velocity of the diver when she hits the water:



According to the equation of kinematics, the velocity of the diver when she hits the water can be determined as follows.



v = v0 - gt

v = 4.00 m/s - 9.8 m/s^2*1.03 s

v = -6.09 m/s

The negative sign indicates that the direction of motion is downward. So,

v = 6.09 m/s

The velocity of the diver when she hits the water is 9.09 m/s.