Answer:
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation: let the following
From the first equation of linear motion
V = u+at..........1
parameters be represented as :
t = Time taken
v = Final velocity
a = Acceleration due to gravity = 9.8m/s²
u = Initial velocity = 4 m/s
s = Displacement
V = 0
Substitute the values into equation 1
0 = 4-9.8(t)
-4 = -9.8t
t = 4/9.8
t = 0.408s
From : s = ut+1/2at^2.........2
S = 4×0.408+0.5(-9.8)×0.408^2
S= 1.632-4.9(0.166)
S = 1.632-0.815
S = 0.817m
Her highest height above the board is 0.817 m
Total height she would fall is 0.817+1.90 = 2.717 m
From equation 2
s = ut+1/2at^2
2.717 m = 0t+0.5(9.8)t^2
2.717 m = 0+4.9t^2
2.717 m = 4.9t^2
2.717/4.9 = t^2
0.554 =t^2
t =√0.554
t = 0.744s
Hence, her feet were in the air for 0.744+0.408seconds
= 1.152s
Also recall from equation 1
V= u+at
V = 0+9.8(0.744)
V = 7.29m/s
Hence, the velocity when she hits the water is 7.29m/s
Finally,
a = 1.152s
b = 0.817 m
c = 7.29m/s
Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.
