(i) The projectile is fired from the top 0 of a hill with speed ux = 98 ms−1 and uy = 0 along the horizontal as shown. It reaches the target P at vertical depth OA, in the coordinate system as shown.
OA = y = 490 m
From second equation of motion,
\(y = u_yt + \frac 12gt^2\)
\(= \frac 12 gt^2\)
⇒ \(490 = \frac 12 \times9.8 t^2\)
⇒ \(t = \sqrt {100} = 10 s\)
(ii) Distance of the target from the hill is given by
AP = x = Horizontal velocity × time
= 98 × 10
= 980 m
(iii) The horizontal and vertical components of velocity v of the projectile at point P are
vx = u = 98 ms−1
vy = uy + gt
= 0 + 9.8 × 10
= 98 ms−1
∴ v = \(\sqrt {{v_x}^2 + {v_y}^2}\)
\(= \sqrt{98^2 + 98^2}\)
\(= 98\sqrt 2\) ms-1