Question

1- A long jumper leaves the ground at an angle of 17.0

Answer 1

1)

height = (u sin theta)^2 / 2*g

=( 10.9*sin 17)^2 / 2*9.81

=0.5176 m

range = u^2 sin (2* theta)/g

= (10.9)2 * sin 34 / 9.81

=6.77 m

2)

The horizontal velocity x time is the range R

R = u cos theta. T

The time to go up and come again down is

T = 2 U sin theta / g.

Eliminating T , R = u^2 sin 2theta / g.

H = usin theta * [T/2] = [U sin theta] ^2 / 2g

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The range R = u^2 sin 2theta / g = 1.2 m

Hence U^2 = 1.2*9.8 / sin 2*42

U = 3.44 m/s.

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The max height reached

H = [U sin theta] ^2 / 2g = [3.44 sin 42] ^2/ (2*9.8)

=0.27 m

3)

vertical component of velocity = u sin theta = 24 sin 30 = 12 m/s

height = -46 m

g = -9.8 m/s^2

a)

h = ut + 0.5 a*t*t

-46 = 12t -4.9t^2

4.9t^2 -12t -46 = 0

t = 4.524 seconds

b)

using conservation of energy

inital energy = ke + pe

=0.5*m*24*24 + m*9.8*46

final energy = kinetic energy = 0.5*m*v*v

equating both

0.5*m*24*24 + m*9.8*46 =0.5* m*v*v

v = 38.44 m/s

c)

horizontal range = horizontal component of velocity * time of flight

= u cos theta * t

=24*cos 30 * 4.524

= 94.23 meters..

4)

since the height is not mentioned im taking the height from problem 3

Let initial speed be U m/s and time of flight be T seconds.

Initial Vertical component velcoity = Usin theta = Usin 28

Horizontal component velocity = Ucos theta = Ucos 28

Height = 46 m and Range = 52.5m .

X = Ucos28 x T

52.5 = Ucos28 x T

46 = 0.5(9.8)(T)^2 + Usin28xT

T = 52.5/Ucos 28

46 = 0.5(9.8)(52.5/Ucos28)^2+ Usin28 /(52.5 /Ucos28)

46 = 0.5(9.8)(52.5/Ucos28)^2+ 52.5 Tan 28

U = 42.89 m/s

T = 1. 39 s

Vertical componenet at impact = 42.89 sin 28 + (1.39 )(9.8) = 33.72 m/s

Speed at impact = sqrt[(42.89cos28)^2 + (33.72)^2 = 50.71 m/s

Angle =Arc tan 33.72/42.89 cos28 = 41.68 degrees below horizontal