Answer:
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
