Question

During a javelin throw event, an athlete throws at a speed of 75 km/hr at an angle of 59 degrees with the horizontal. Determine the maximum height, time of flight and horizontal distance of the javelin. Use the equation for projectile motion and constant acceleration. 

Answer 1

Explanations & Calculations.

$\qquad\qquad \begin{aligned} \small 75\,kmh^{-1}&=\small \frac{75\times1000\,m}{3600\,s}\\ &=\small 20.8\,ms^{-1} \end{aligned}$

• To calculate the maximum height it will reach, apply $\small v^2=u^2+2as$ for its vertical motion. (in a projectile motion, an object has motion in both vertical & horizontal directions)

$\qquad\qquad \begin{aligned} \small v^2&=\small u^2+2as\\ \small 0^2 &=\small (20.8\sin59\,ms^{-1})^2+2(-9.8\,ms^{-2})\times h\\ \small h&=\small \bold{16.2\,m} \end{aligned}$

• To calculate the time of flight, apply $\small s=ut+\frac{1}{2}at^2$ for the projectile's vertical motion.

$\qquad\qquad \begin{aligned} \small 0\,m&=\small (20.8\sin59)t+0.5\times(-9.8\,ms^{-2})t^2\\ \small t&=\small 0\,s\qquad\text{or}\qquad 3.6\,s\\ \small\therefore\,\:t&=\small \bold{3.6\,s} \end{aligned}$

• To calculate the horizontal distance, apply $\small s=ut+\frac{1}{2}at^2$ to its horizontal motion.
• For the horizontal motion, the acceleration is zero this equation simplifies as follows and the calculations become easy.

$\qquad\qquad \begin{aligned} \to \\ \small s&=\small (20.8\cos59\,ms^{-1})\times3.6\,s\\ &=\small \bold{38.6\,m} \end{aligned}$