Range of both the particles is R = u^2 sin 2 thita /g = (49)^2 sin 90 degree/9.8 =245 m Range of both the particles is R = u^2 sin 2 thita /g = (49)^2 sin 90 degree/9.8 =245 m February 13, 2021 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) , Facebook Messenger WhatsApp Share this:TwitterFacebook Related
