Question

During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16m/s. While making the throw, his hand pushed the shot a distance of 1.6m. Assume the acceleration was constant during the throw. A) Determine the acceleration of the shot.

B) Determine the time it takes to accelerate the shot.
C) Determine the horizontal component of the force exerted on the shot by hand.

Answer 1

Step-by-step 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.

Answer 2

(a) Acceleration of the shot put is 80
`m/s^(2)`

(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,
`a_(s)` is given by eqn 2 of motion:

`v'^(2) = v^(2) + 2a_(s)d`

`16^(2) = 0 + 2a_(s)* 1.6`

`a_(s) = 80 m/s^(2)`

(b) Time taken, t is given by eqn 1 of motion:

v' = v +
`a_(s)t`

16 = 0 + 80t

t = 0.2 s

(c) Horizontal component of force is given by:

`F = ma_(s)`

where

m = mass = 7.9 kg

Now,

`F = 7.9* 80 = 632 N`