Respuesta :
Answer:
(a) Acceleration of the shot put is 80[tex]m/s^{2}[/tex]
(b) Time taken for accelerating the shot is 0.2 s
(c) Horizontal force is 632 N
Solution:
As per the question:
Final speed of the shot put throw, v' = 16 m/s
Initial speed, v = 0 m/s
The distance covered by the shot put, d = 1.6 m
The shot put throw was at constant acceleration
Now,
(a) Acceleration of the shot put, [tex]a_{s}[/tex] is given by eqn 2 of motion:
[tex]v'^{2} = v^{2} + 2a_{s}d[/tex]
[tex]16^{2} = 0 + 2a_{s}\times 1.6[/tex]
[tex]a_{s} = 80 m/s^{2}[/tex]
(b) Time taken, t is given by eqn 1 of motion:
v' = v + [tex]a_{s}t[/tex]
16 = 0 + 80t
t = 0.2 s
(c) Horizontal component of force is given by:
[tex]F = ma_{s}[/tex]
where
m = mass = 7.9 kg
Now,
[tex]F = 7.9\times 80 = 632 N[/tex]
Answer:
Explanation:
The force acts on the throw during which hand moved by 1.6 m
So it is a case of uniformly accelerated motion
s = 1.6 m
u = 0
v = 16 m /s
v² = u² + 2 a s
16 x 16 = 0 +2 x a x 1.6
a = 80 m / s²
v = u + at
16 = 0 + at
16 = 80 t
t = 1 /5 = .2 s.
force = mass x acceleration
= 7.9 x 80
= 632 N.
