Question

A world-class shotputter can put a 7.26 kg shot a distance of 22 m. Assume that the shot is constantly accelerated over a distance of 2 m at an angle of 45 degrees and is released from a height of 2 m above the ground. Estimate the weight that this athlete can lift with one hand.

Answer 1

To solve this problem we will make a free body diagram to better understand the displacement measurements made by the body. From there we will apply the linear motion kinematic equations that describe the position of the body in reference to its vertical displacement, acceleration and speed. With this speed found we will apply the energy conservation theorem that will allow us to find the Force.

Equation of trajectory of a projectile is

$y = xtan\theta - x^2 \frac{g}{2u^2cos^2\theta}$

Here

u = Initial velocity

x = Horizontal displacement

g = Acceleration due to gravity

y = Vertical displacement

We have that

$x = 22m$

$y = -2\sqrt{2m}$

Replacing we have that,

$-2\sqrt{2} = 22tan45\° -\frac{9.8}{2u^2cos^2 45}(22)^2$

$-2\sqrt{2} =22 -\frac{9.8*484}{2u^2(1/2)}$

$-2\sqrt{2} =22 -\frac{4743.2}{u^2}$

$u^2 = \frac{4743.2}{22+2\sqrt{2}}$

$u = 191.03m/s$

From the work energy theorem

$W_{net} = K_f +K_i$

Here,

$K_f = \frac{1}{2} mu^2$

$K_i = \frac{1}{2} m(0)^2 = 0$

$W_{net} = W_m+W_g$

Where,

$W_m =\text{Work by man} = F_s$

$W_{g} = \text{Work by gravity} = -mgh$

Therefore

$F_s -mgh = \frac{1}{2} mu^2$

$F_s = \frac{m}{s} (\frac{u^2}{2}+gh)$

$F_s = \frac{7.26}{2\sqrt{2}}(\frac{191.03}{2}+9.8*2)$

$F_s = 295.477N$