Kinematics in Physics please help me

Question

Miss Loren Chavez was with her fiancé on the top of 100 m building when she suddenly get mood swings and get mad at him. So she impulsively kicked him at 16 ft/s angle of <15 off the ground. Determine her fiancé's :

a) Maximum elevation

b) Total air flight time

c) Horizontal displacement

d) Velocity as he touch the ground

And please can kindly show me your formula with unit of measures and kindly solve this thing in ft units. Thank You Very Much!!!

Answer 1

given:

H = 100 m

v = 16 ft/s

angle A = 15 deg

Determine

a) Maximum elevation

b) Total air flight time

c) Horizontal displacement

d) Velocity as he touch the ground

Solution:

Units should be in feet

H = 100 m * 3.28 ft/m

Ht of building

H = 328 ft

The motion in x and y direction is independent, so compute for the horizontal(x) and vertical (y) components of the velocity.

Vox = Vo cos15 = 16*0.966 = 15.46 ft/s

Voy = Vo sin15 = 16*0.259 = 4.14 ft/s

a) Assuming negligible air resistance, the motion on the y direction is free fall and is affected by gravity

Vf^2 - Vo^2 = -2g ymax

at ymax or at the peak of the flight, Vfy = 0

Vfy^2-Voy^2 = -2g (ymax - yo)

-Voy^2 = -2g (ymax - yo)

-(4.14)^2 = -2(32.2) (ymax -328)

ymax -328 = -(4.14)^2/(-64.4)

ymax = -17.2/-64.4 + 328

ymax = 328.27 ft

b) Δy = Voy t - 0.5g t^2

yf - yo = Voy t - 0.5g t^2

0 - 328 = 4.14 t - 16.1 t^2

16.1 t^2 - 4.14 t -328 = 0

t^2 - 0.26t = 20.4

t^2 - 0.26t + 0.13^2= 20.4 +0.13^2

(t - 0.13)^2 = 20.4 + 0.0169

t - 0.13 = sqrt(20.4169)

t = 4.52 + 0.13

t = 4.65 sec

c) Horizontal displacement

Mtion in the horiz direction is constant

x = Vox t

x = 15.46 (4.65)

x = 71.9 ft

d) Velocity as he touch the ground, y=0

Vfy^2 - Voy^2 = -2g (yf-yo)

Vfy^2 - 4.14^2 = -2(32.2)(0-328)

Vfy^2= 21123 + 17

Vfy^2= 21140

Vfy = sqrt(21140)

Vfy = -145.4 ft/s (choose the negative since the direction is downwards)

Vfx = 15.46 ft/s = constant

the resultant velocity is the vector sum of Vfx nad Vfy (use pythagorean theorem)

Vf^2 =  Vfx^2 + Vfy^2

Vf^2 =  (15.46)^2 + (145.4)^2

Vf =sqrt(21380.2)

Vf = 146.2 ft/s

angle = arctan (Vfy/Vfx)

angle = arctan (-145.4/15.46)

angle = arctan (-9.405)

angle = -84 degrees ( negative sign means the direction is downwards)

Answer 2

First of all, let us be consistent with the units. The building height is given in METERS (100 m) while the initial velocity is given in FT/SEC (16 ft/sec).

For this particular problem, you specified the solution to in &quot;ft&quot; units. First step is to convert the building height of 100 meters to its equivalent in feet.

Conversion factor : 3.28 feet = 1 meter

100 meters = 100 * 3.28 = 328 ft.



&lt;&lt; a) Maximum elevation &gt;&gt;

Maximum elevation is given by the formula

Y = V^2(sin A)/(2g) + 328

where

Y = maximum elevation as measured from the ground

V = initial velocity = 16 ft/sec (given)

A = angle of launch = 15 degrees (given)

g = acceleration due to gravity = 32.2 ft/sec^2 (constant)

Substituting appropriate values,

Y = 16^2(sin 15)/(2 * 32.2) + 328

Y = 0.27 + 328 = 328.27

Y = 328.27 feet

&lt;&lt; b) Total air flight time &gt;&gt;

There are three components of the flight time:

1. Time to reach maximum height (as measured from the top of the building)

2. Time to go down to the same elevation as the building

3. Time to descend from 100 meters (328 feet) high to the ground below

Let

T1 = time to reach maximum height (as measured from the top of the building)

T2 = Time to go down to the same elevation as the building

T3 = Time to descend from 100 meters (328 feet) high to the ground below

NOTE that T1 = T2, hence total flight time

T = 2(T1) + T3

The formula to determine T1 is given by

T1 = V(sin A)/g

where all terms have been previously defined. Therefore,

T1 = 16(sin 15)/32.2

T1 = 0.13 sec

To solve for T3, the following formula is used

H = VyT3 + (1/2)gT3^2

where

H = 328 feet

Vy = vertical component of the velocity = 16(sin 15)

and all the other terms have been previously defined.

Substituting values,

328 = 16(sin 15)(T3) + (1/2)(32.2)T3^2

328 = 4.14(T3) + 16.1(T3^2)

Rearranging the above,

16.1(T3^2) + 4.14(T3) - 328 = 0

The above is a quadratic equation and using the quadratic formula,

T3 = 7.17 sec.

Therefore, total flight time,

T = 2(0.13) + 7.17  

T = 7.43 sec.

&lt;&lt; c) Horizontal displacement &gt;&gt;

X = horizontal displacement = Vx(T1 + T2) + Vx(T3) = Vx(T1 + T2 + T3)

and since (T1 + T2 + T3) = total flight time = 7.43

where

Vx = horizontal component of the velocity = 16(cos 15)

Modifying the above,

X = Vx(T1 + T2 + T3)

and since (T1 + T2 + T3) = total flight time = 7.43

Substituting values,

X = 16(cos 15)(7.43)

X =  114.83 feet (measured from the base of the building)

d) Velocity as he touch the ground



This is given by the formula

Vg - Vy = g(T3)

where

Vg = velocity as he hits the ground

Vy = vertical component of velocity = 16*sin 15

g = 32.2 ft/sec^2

T3 = 7.17 (as calculated above)

Substituting appropriate values,

Vg - 16(sin 15) = (32.2)(7.17)

Solving for Vg,

Vg = 16(sin 15) + 32.2(7.17)

Vg = 235.02 ft/sec