Question

2d kinematics problem
in a shot - put event, an athlete throws the shot with an initial speed of 12.0 m/s at an angle of 40.0 degrees above the horizontal. the shot leaves their hand at a height of 1.80 m. how far does the shot travel horizontally before it hits the ground?
1d kinematic equations for constant acceleration
$v_f = v_i+at$
$v_f^2 = v_i^2 + 2adelta x$
$delta x=v_it+\frac{1}{2}at^2$
$delta x=\frac{v_i + v_f}{2}t$

Explanation:

Step1: Resolve initial velocity

$v_{0x}=v_0\cos\theta = 12\cos40^{\circ}\approx9.19\ m/s$
$v_{0y}=v_0\sin\theta = 12\sin40^{\circ}\approx7.71\ m/s$

Step2: Use vertical - motion equation

The vertical - displacement equation is $\Delta y = v_{0y}t-\frac{1}{2}gt^{2}$, where $\Delta y=- 1.80\ m$ (downward), $g = 9.8\ m/s^{2}$.
So, $-1.80=7.71t-4.9t^{2}$.
Rearranging gives $4.9t^{2}-7.71t - 1.80 = 0$.

Step3: Solve quadratic equation for time

Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 4.9$, $b=-7.71$, $c=-1.80$.
$t=\frac{7.71\pm\sqrt{(-7.71)^{2}-4\times4.9\times(-1.80)}}{2\times4.9}=\frac{7.71\pm\sqrt{59.44 + 35.28}}{9.8}=\frac{7.71\pm\sqrt{94.72}}{9.8}=\frac{7.71\pm9.73}{9.8}$.
We take the positive root $t=\frac{7.71 + 9.73}{9.8}=\frac{17.44}{9.8}\approx1.78\ s$.

Step4: Calculate horizontal distance

Using the horizontal - motion equation $x = v_{0x}t$.
$x=9.19\times1.78\approx16.4\ m$.

Answer:

$16.4\ m$