Question

A stone is thrown with an initial velocity components of 20 m/s along the vertical, and 15 m/s along the horizontal direction. Determine the position and velocity of the stone after 3 s. Determine the maximum height that it will reach and the total distance travelled along the horizontal on reaching the ground. (Assume g = 10 m/s²)

Answer 1

The initial velocity of the stone in x-direction = u cos θ = 15 m/s and in y-direction = u sin θ = 20 m/s

After 3 s, v_x = u cos θ = 15 m/s

v_y = u sin θ – gt

= 20 – 10(3)

= -10 m/s

10 m/s downwards.

∴ v =

∴ v = 18.03m/s

tan α = v_y/v_x = 10/15 = 2/3

∴ α = tan^-1 (2/3) = 33° 41’ with the horizontal.

S_x = (u cos θ)t = 15 × 3 = 45m,

S_y = (u sin θ)t – \(\frac{1}{2}\)gt² = 20 × 3 – 5(3)²

∴ S_y = 15m

The maximum vertical distance travelled is given by,

H = \(\frac{(u\,sin\,\theta)^2}{2g}=\frac{20^2}{(2\times10)}\)

∴ H = 20m

Maximum horizontal distance travelled