Question

In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? (although the maximum distance for a projectile on level ground is achieved at 45º when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38º will give a longer range than 45º in the shot put.)

Answer 1

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance R horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

$R=u_xt=(ucos \theta)t$

Therefore,

$t=\frac{R}{ucos\theta} .......(1)$

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

$y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2$

Substitute the value of t from equation (1).

$y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )$

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

$y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s$

The shot put was thrown with a speed 15.02 m/s.