S Smilebuffalo Member Joined Jun 9, 2008 Messages 89 Location Fairfield West Gender Male HSC 2010 Aug 27, 2009 #1 An object is projected with horizontal velocity 8ms^-1 and vertical velocity 5ms^-1. Find the range of its flight. (use g = 10ms^-2)
An object is projected with horizontal velocity 8ms^-1 and vertical velocity 5ms^-1. Find the range of its flight. (use g = 10ms^-2)
D davidgoes4wce Well-Known Member Joined Jun 29, 2014 Messages 1,877 Location Sydney, New South Wales Gender Male HSC N/A May 8, 2015 #2 Vertical Component ÿ=-10 ẏ=∫ -10 dt= -10t + C t=0 y=5 C=5 ẏ=-10t + 5 y=∫-10t+5 dt y=-5t^2+5t+C t=0,y=0 C=0 y=-5t^2+5t let y=0 0=-5t(t-1) t=1 Horizontal Components ẍ=0 ẋ=∫ 0 dt= 0t +C when t=0, ẋ=8 ẋ=8 x=∫8 dt when t=0, x=0 x=8t+ C C=0 x=8t let t=1 x=8*1=8 metres
Vertical Component ÿ=-10 ẏ=∫ -10 dt= -10t + C t=0 y=5 C=5 ẏ=-10t + 5 y=∫-10t+5 dt y=-5t^2+5t+C t=0,y=0 C=0 y=-5t^2+5t let y=0 0=-5t(t-1) t=1 Horizontal Components ẍ=0 ẋ=∫ 0 dt= 0t +C when t=0, ẋ=8 ẋ=8 x=∫8 dt when t=0, x=0 x=8t+ C C=0 x=8t let t=1 x=8*1=8 metres
M madanu New Member Joined May 20, 2024 Messages 1 Gender Male HSC 2024 May 20, 2024 #3 what happened to the sin
KloppsAndRobbers Active Member Joined Oct 17, 2023 Messages 111 Gender Male HSC 2024 May 20, 2024 #4 madanu said: what happened to the sin Click to expand... There's no need for sin or cos because the velocities are already given.
madanu said: what happened to the sin Click to expand... There's no need for sin or cos because the velocities are already given.