Ineed with #25 and #26 i don't understand it .. can somebody me with it
it is due tomorrow i have to get it done by tomorrow morning if not i will get a f on it

25.

The question is on  kinematic equations and free fall

An expression for h, displacement is given by;

h=v.t +1/2 at²................... where v is initial velocity, t is time and a is acceleration due to gravity

Given v=10ft/s d=-50 and a= -32.2 ft/s²..............................substitute the values in

h=v.t +1/2 at².

-50=10t + 1/2( -32.2)t²

-50=10t-16.1t²

16.1t²-10t-50=0..........................the function for height h in feet

b)  Solve 16.1t²-10t-50=0 using the quadratic formula

t= (-b ± √b² - 4ac )/2a

a=16.1  b= -10   and c= -50

t=( 10 ± √ (-10)² - 4 × 16.1 × -50 ) / 2×16.1

t= (10± √3320 )/ 32.2

t= (10± 57.62 ) / 32.2

t=67.62/32.2 = 2.1 sec or  - 47.62/32.2 = -1.5 sec

26.

a)The question is on  kinematic equations and free fall

An expression for h, displacement is given by;

h=v.t +1/2 at²................... where v is initial velocity, t is time and a is acceleration due to gravity

Given v=3ft/s  h= -1.3 ft and a= -32.2 ft/s²

substitute vales in;

h=v.t +1/2 at²

1.3=3t+1/2(32.2)t²

1.3=3t+16.1t²

16.1t²+3t-1.3=0...........................the function for h

b) Solve for t in 16.1t²+3t-1.3=0 using the quadratic formula

a=16.1        b=3      c= -1.3

t= (-3 ± √ -3² - 4×16.1× -1.3) / 2×16.1

t= ( -3 ± √ 9 + 83.72 ) / 32.2

t= (-3 ± 9.6 )/32.2

t= (-3+9.6)/32.2  = 6.6/32.2 = 0.2049 or

t= (-3-9.6)/32.2 = -12.6/ 32.2 = -0.391

c)  if the ball hit the rim at one half foot above the ground, find the distance it covered before hitting the rim

1.3-0.5=0.8 ft......................displacement

applying h=v.t +1/2 at²

0.8=3t+16.1t²

16.1t²+3t-0.8=0.......................solve using the quadratic formula

a=16.1  b=3    c= -0.8

t=(  -3 ± √3²- 4×16.1×-0.8  )/2×16.1

t=( -3±√9+51.52 )/ 2× 16.1

t = (-3 ± √60.52 )/32.2

t=( -3±7.78 )/ 32.2

t= (-3+7.78 )/32.2 =0.148 sec   or  t= (-3-7.78)/32.2 = -0.335 sec

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