A stone was thrown with velocity 20m/s at an angle of 30 degrees from a height h. The stone moves under gravity freely and reaches the floor 5s after thrown. a) Find H, b)the horizontal distance covered

To find the height we first need to find the initial vertical velocity of the stone. This will be equal to 20 * sin(30) = 200.5 = 10m/s. From here we can use SUVAT. In this case s=-h,u=10m/s, v is unknown, a = -9.8m/s^2 (g), t = 5s. So as s = ut + 0.5at^2, so -h = 105 + -4.95^2 = 50 - 122.5 = -72.5. So h = 72.5m. Now we find the horizontal distance. To do this we first need the initial horizontal velocity. This is 20cos(30) . Now we use SUVAT again. s = ut + 1/2 at^2, as the acceleration in a horizontal direction is 0, we can just use s = u*t, so s = 20cos(30) * 5 = 100cos(30) = 86.6m.