A stone is thrown from a bridge 10m above water at 30ms^-1 30 degrees above the horizontal. How long does the stone take to strike the water? What is its horizontal displacement at this time?

The first thing to do for part a) is work out the vertical component of the velocity. By drawing a triangle where 30m/s is the hypotenuse with angles of 30 degrees, 90 degrees, and 60 degrees, and by applying the rule that sin(theta) = opposite/hypotenuse we can see that the vertical velocity will be 30sin30 which is 30 * 1/2 = 15 We can then use the SUVAT s=ut+1/2at^2. Substituting in our values for s (-10m), u (15ms^-1) and a (-g which is -9.8ms^-2) we can obtain a quadratic equation, solveable for t. -10 = 15t - 4.9t^2 -> 4.9t^2-15t-10 = 0 . Is is important that s and a have negative signs as our reference direction is upwards. We can either solve this using a graphical calculator, or by using the quadratic formula. This gives us our answer of (15+sqrt(15^2+44.910))/24.9 =3.62s (3sf). We knew to add the sqrt part of the equation because subtracting it would have given us a negative time which is impossible. We can check our answer by inputting our t into the equation and checking our s output. b) As we know t, we simply use distance=speedxtime to calculate the stone's horizontal displacement. This is because gravity acts purely downwards so the stone cannot accelerate horizontally. Speed = 30cos30 = 26.0 (3sf) which we know from our triangle earlier, so the displacement = 94.1m (3sf).