Given, initial speed of particle,
v=(3)m∕s Acceleration of particle,
a=(−1−0.5)m∕s2 As we know that,
sx=vxt+axt2. . . (i)
and
sy=vyt+ayt2 . . . (ii)
where,
vx,vy is the initial speed of particle along
X,Y-axes respectively,
t be the time taken by the particle to reach maximum height.
∴sx=3t−(1)t2 On differentiating both sides w.r.t
t, we get
=3−t=vx′ where,
vx′ is final speed of particle along
X-axis.
∵ At maximum distance\/ height, speed becomes
0ms−1 ⇒t=3s ∴sx=3×3−(1)×32 =9−=m sx=m and
sy=uyt−ayt2 ⇒sy=0×3−××32 =m ∴sy=−() ∵ Particle reaches to maximum height and returns.
∴sy=−×2 =− ∴s=sx+sy=− =(−)
