Refresher - MATHEMATICS Quiz 16
PROBLEM 1:
A tennis ball is thrown vertically downward from a height of 54 ft with an initial velocity of 8
fps. What Is the impact velocity of it hits a 6 ft. tall person on the head?
Solution:
v(t) = ∫ a(t)dt
V1= 8 fps
v(t) = - 32 t + c1
when v 0 = - 8
t=0
- 8 = c1
c1 = - 8 48
54
v(t) = - 32t - 8
s(t) = ∫ ( - 32t - 8)dt V2= ?
s(t) = - 16t 2 - 8t + c 2
6’
When s = 54
t=0
54 = - 16t 2 - 8t + c 2
54 = 0 + c 2
c 2 = 54
when s = 6
6 = - 16t 2 - 8t = 54
v(t) = - 32 + c1
2
16t + 8t - 48 = 0
32(3)
8(2t 2 + t - 6) = 0 v(t) = - -8
2
(2t 2 - 3)(t + 2) = 0
v(t) = - 56 fps.
t = 3/2 sec.
, Refresher - MATHEMATICS Quiz 16
PROBLEM 2:
In triangle ABC, AB = 30 m., BC = 36 m. and AC = 48 m. has an area
equal to 539.32 m2. How far is the point of intersection of the
perpendicular bisectors from A?
Solution:
B
30 r
36
r r
A
48 C
abc
A=
4r
(36)(48)(30)
539.32 =
4r
r = 24.03 m. from A
PROBLEM 1:
A tennis ball is thrown vertically downward from a height of 54 ft with an initial velocity of 8
fps. What Is the impact velocity of it hits a 6 ft. tall person on the head?
Solution:
v(t) = ∫ a(t)dt
V1= 8 fps
v(t) = - 32 t + c1
when v 0 = - 8
t=0
- 8 = c1
c1 = - 8 48
54
v(t) = - 32t - 8
s(t) = ∫ ( - 32t - 8)dt V2= ?
s(t) = - 16t 2 - 8t + c 2
6’
When s = 54
t=0
54 = - 16t 2 - 8t + c 2
54 = 0 + c 2
c 2 = 54
when s = 6
6 = - 16t 2 - 8t = 54
v(t) = - 32 + c1
2
16t + 8t - 48 = 0
32(3)
8(2t 2 + t - 6) = 0 v(t) = - -8
2
(2t 2 - 3)(t + 2) = 0
v(t) = - 56 fps.
t = 3/2 sec.
, Refresher - MATHEMATICS Quiz 16
PROBLEM 2:
In triangle ABC, AB = 30 m., BC = 36 m. and AC = 48 m. has an area
equal to 539.32 m2. How far is the point of intersection of the
perpendicular bisectors from A?
Solution:
B
30 r
36
r r
A
48 C
abc
A=
4r
(36)(48)(30)
539.32 =
4r
r = 24.03 m. from A
