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//s^2` )

Explanation

The initial velocity of the stone in x-direction`=u cos theta = 15 m//s " and in y-direction " = u sin theta = 20 m//s`.
After `3 s, v_x = u cos theta = 15 m//s and v_y = u sin theta - "gt" = 20 - 10(3)= - 10 m//s = 10 m/s` downwards.
`:. v= sqrt(v_x^2 + v_y^2) = sqrt(15^2 +10^2)`
`= sqrt(225+100) = sqrt325 = 18.03 m//s`
`tan alpha =v_y //v_x = 10//15 = 2//3`
`:. alpha = tan^(-1) (2//3) = 33^@ 41.` with the horizontal. `S_x= (u cos theta)t = 15xx3 = 45 m`
`s_y = (u sin theta) t -1/2 "gt"^2 = 20 xx3 - 5(3)^2 = 15 m`.
Thus the stone will be at a distance 45 m along horizontal and 15 m along vertical direction from the initial position after time 3 s. The velocity is 18.03 m/s making an angle `33^@ 41` with the horizontal. The maximum vertical distance travelled is given by `H=(u sin theta)^2//(2g) = 20^2//(2 xx10) = 20 m`
Maximum horizontal distance travelled
`R = 2. u_x .u_y //g=2(15)(20)//10=60m`