Answer to Question #256745 in Mechanics | Relativity for kkk

Question #256745

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. 


1
Expert's answer
2021-10-26T17:18:51-0400

Explanations & Calculations.


75kmh1=75×1000m3600s=20.8ms1\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 v2=u2+2as\small v^2=u^2+2as for its vertical motion. (in a projectile motion, an object has motion in both vertical & horizontal directions)

v2=u2+2as02=(20.8sin59ms1)2+2(9.8ms2)×hh=16.2m\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 s=ut+12at2\small s=ut+\frac{1}{2}at^2 for the projectile's vertical motion.

0m=(20.8sin59)t+0.5×(9.8ms2)t2t=0sor3.6s  t=3.6s\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 s=ut+12at2\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.

s=(20.8cos59ms1)×3.6s=38.6m\qquad\qquad \begin{aligned} \to \\ \small s&=\small (20.8\cos59\,ms^{-1})\times3.6\,s\\ &=\small \bold{38.6\,m} \end{aligned}




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