Respuesta :
Answer:
a) 12.528 m/s
b) 15.344 m/s
Explanation:
Given:
Mass of the child, m = 34.9 kg
Height of the water slide, h = 16.0 m
Now,
a) By the conservation of energy,
loss in potential energy = gain in kinetic energy
mgh = [tex]\frac{\textup{1}}{\textup{2}}\textup{m}\times\textup{v}^2[/tex]
where,
g is the acceleration due to the gravity
v is the velocity of the child
thus,
at halfway down, h = [tex]\frac{\textup{16}}{\textup{2}}[/tex]= 8 m
therefore,
34.9 × 9.81 × 8 = [tex]\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2[/tex]
or
v = 12.528 m/s
b)
at three-fourth way down
height = [tex]\frac{\textup{3}}{\textup{4}}\times16[/tex] = 12 m
thus,
loss in potential energy = gain in kinetic energy
34.9 × 9.81 × 12 = [tex]\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2[/tex]
or
v = 15.344 m/s
Answer:
(a) 12.52 m/s
(b) 15.34 m/s
Explanation:
mass, m = 34.9 kg
h = 16 m
(a) Initial velocity, u = 0
height = h / 2 = 16 / 2 = - 8 m (downward)
let the speed of child is v.
acceleration, a = - 9.8 m/s^2 (downward)
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=0^{2}+2\times 9.8\times 8[/tex]
v = 12.52 m/s
(b) Initial velocity, u = 0
height = 3 h / 4 = 12 m = - 12 m (downward)
let the speed of child is v.
acceleration, a = - 9.8 m/s^2 (downward)
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=0^{2}+2\times 9.8\times 12[/tex]
v = 15.34 m/s
