Correct Answer - B::C
At t = 0 `v_T = (10hatj)m//s`
`v_(ST) = 10 cos 37^@hat k - 10 sin 37^@hati = (8hatk-6hati)m//s`
`:. V_s = v_(ST) + v_(T) = (-6hati + 10hatj+8hatk)m//s`
(a) At highest point vertical component `(hatk)`of `v_s` will become zero. Hence, velocity of particle at highest point will become `(-6hati+10hatj)m//s`.
(b) Time of flight, `T = (2v_Z)/g = (2xx8)/10 = 1.6s`
`x = x_i + v_xT`
`= 16/pi -6 xx 1.6 = -4.5m`
`y = (10)(1.6) = 16 m and z = 0`
Therefore , coordinates of particle where it finally
lands on the ground are `(-4.5m, 16m, 0)`
At highest point, `t = T/2 = 0.8s`
`:. x = 16/pi - (6)(0.8) = 0.3m`
`y = (10)(0.8)= 8.0m`
and `z = (v_(Z)^(2))/(2g) = (8)^2/20 = 3.2m`
Therefore , coordinates at highest point are,
`(0.3 m, 8.0m, 3.2m)`.