Trajectory of particle in a projectile motion is given as \( y=x-x^{2} / 80 \). Here, \( x \) an...
Channel:
Subscribers:
443,000
Published on ● Video Link: https://www.youtube.com/watch?v=FXsjgd_W6fg
0Trajectory of particle in a projectile motion is given as \( y=x-x^{2} / 80 \). Here, \( x \) and \( y \) are in metres and considered along horizontal and vertical direction respectively \( \left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right) \). For this projectile motion.
(a) angle of projection is \( 45^{\circ} \)
(b) angle of velocity with horizontal after \( 4 \mathrm{~s} \) is \( \tan ^{-1}(1 / 2) \)
(c) maximum height is \( 80 \mathrm{~m} \)
(d) horizontal range is \( 20 \mathrm{~m} \)
📲PW App Link - https://bit.ly/YTAI_PWAP
🌐PW Website - https://www.pw.live