Physics HARD PROBLEM?

Question

A high diver leaves the end of a 5.0 m high diving board and strikes the water 1.4 s later, 2.2 m beyond the end of the board. Considering the diver as a particle, determine the following.
(
a) her initial velocity, v0
____ m/s ____° above the horizon

(b) the maximum height above the water reached
______ m
(c) the velocity vf with which she enters the water
_____ m/s _____° below the horizon

Answer 1

Looks like you've got plenty of physics assignment there dude! Lemme see if I can do this one.
This is a problem in projectile...

Given:
5.0 m = height of diving board
1.4 sec = time elapsed before the diver strikes
the water
2.2 m = horizontal distance reached by diver from
the end of the board (the diver left the
end of the diving board)
Find:
a) Vi = initial speed (m/sec) of diver
A = angle above the horizontal
b) Dmax = max. height (m ) above the water
reached by diver
c) Vf = speed she enters the water (m/sec)
B = angle below the horizontal.

Solution:
a) Let Vi = initial speed of the diver
Then, Viy = vertical component of Vi
= ViSin(A),
Vix = horizontal component of Vi
= ViCos(A)
Using the following formula for free-falling body:
Choosing downward as positive and upward as negative:
Dy = (Viy)t + (1/2)gt^2
- 5 = ViSin(A)(1.4 sec) + (1/2)(9.8m/s^2)(1.4^2sec^2)

- 5 = ViSin(A)(1.4) + 9.604Eq.1

Dx = ViCos(A)(t)
2.2 = ViCos(A)(1.4) Eq. 2

Arrange Eq. 1:
ViSin(A)(1.4) = -5 - 9.604
ViSin(A) = (-5 - 9.604)/1.4
ViSin(A) = 3.289
Arrange Eq. 2:
ViCos(A) = 2.2/1.4
ViCos(A) = 1.571
Divide: Eq.1/Eq.2:
ViSin(A)/ViCos(A) = 3.289/1.571
Tan(A) = 2.094
A = arctan(2.094)
A = 64.5 deg above the horizontal
Therefore, from Eq. 2:
Vi(Cos 64.5 deg) = 1.571
Vi = 1.571/Cos 64.5 deg.
Vi = 3.6 m/sec
ANS: 3.6 m/sec 64.5 deg. above the horizontal

b) Let Dmax = max. height above the water reached.
Dmax = - 5.0 m + dymax, where
dymax = (Vfy^2 - Viy^2)/2g
dymax = (0 - {ViSin(A)}^2/(2x9.8)
= - (3.6Sin 64.5deg)^2/19.6
= - 0.54 m
(Note: I got negative because I chose upward as negative and downwad as positive. See beginning of solution for number 1). Therefore,
Dmax = - 5.0 m + dymax
Dmax = - 5.0 m + - 0.54 m
= - 5.54 m ANS.

c) Vfx = final horizontal velocity = initial horizontal
velocity, which is, Vix
Vfx = Vix = ViCosA
Vfx = 3.6Cos 64.5 deg
Vfx = 1.55 m/sec

Solving for Vfy = final vertical velocity.
Using:
2gDy = Vfy^2 - Viy^2
Vfy = sqrt[2gDy + Viy^2
Vfy = sqrt[2(9.8)(5) + (ViSinA)^2]
Vfy = sqrt[2(9.8)(5) + (3.6Sin 64.5deg)^2]
Vfy = 10.4 m/sec.

Vf = sqrt[Vfy^2 + Vfx^2]
Vf = sqrt[10.4^2 + 1.55^2]
Vf = 10.5 m/sec

Solving for angle B below the horizon:
B = arctan[Vfy/Vfx]
B = arctan[10.4/1.55]
B = 81.5 deg below the horizontal.
Therefore,
Vf = 10.5 m/sec 81.5 deg below the horizontal
ANS.

Hope I help you.

teddyboy

Answer 2

Vx = (2.2 m) / (1.4 s) ≈ 1.571429 m/s
s = s0 + V0t + (1/2)at^2)
0 = 5 + 1.4V0 - (1/2)(9.80665)(1.4^2)
1.4V0 = 9.610517 - 5 = 4.610517
V0y ≈ 3.293226 m/s
V0 ≈ 3.648935 m/s @ 64.5°

h = s0 + (V^2 - V0^2)/(2a)
h = 5 - 3.293226^2)/(2*(- 9.80665))
h ≈ 5.552958 m

Vfx = √(2gh)
Vfx = √(2*9.80665*5.552958)
Vfx ≈ 10.43608 m/s
Vf ≈ 10.97236 m/s @ 72°

Answer 3

a) v = v[0] - 9.8 t
s{y] = v[0y]t - 1/2(9.8)t^2
- 5 = v[0y] (1.4) - (1/2)(9.8)(1.4^2)
-5 = 1.4 v[0y] - 9.6
v[0y] = (9.6 - 5)/1.4 = 3.286 m/s, vertical component
s[x] = v[ox] t
2.2 = v[0x] (1.4)
v[0x] = 1.57 m/s , horizontal component
lv[0]l = sq rt ( 3.286^2 + 1.57^2) = 3.642 m/s with arc tan(3.286/1.57) = 64.5 deg with horizontal

b) h above the water
v = 3.286 - 9.8t = 0
t = 3.286/9.8 = 0.3353 sec
h = 3.286(.3353) - (1/2)(9.8)(.3353^2) = 0.551 m

c) v[fy} = 3.286 - 9.8(1.4) = -10.434 m/s (downward)
v[fx] = 1.57 m/s
lv[f]l = sq rt( 10.434^2 + 1.57^2) = 10.552 m/s with angle arc tan(10.434/1.57) = 81.44 deg below horizontal