Question

A projectile is fired horizontal with a velocity of 98m / s from the top of a hill 490 m high. Find 

(a) the time taken by the projectile to reach the ground

(b) the distance of the point where the particle hits the ground from foot of the hill and

(c) the velocity with which the projectile hits the ground (g= 9.8 m/s²)

Answer 1

(i) The projectile is fired from the top 0 of a hill with speed u_x = 98 ms⁻¹ and u_y = 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

v_x = u = 98 ms⁻¹

v_y = u_y + 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

Answer 2

Here, u_x = 98 m/s, a_x = 0, u_y = 0 and a_y = g

(a) At A, s_y = 490 m. So, applying

or BA = (98) (10) + (0) or BA = 980 m

(c) v_x = u_x = 98 m/s

v_y = u_y + a_y t = 0 + (9.8) (10) = 98 m/s

Thus, the projectile hits the ground with a velocity 98 √2 m /s at an angle of β = 45º with horizontal.