A ball is fired at an angle of 50 degrees with a velocity of 10 ms^-1, at what time does it first hit the floor?

First we must decide how we can begin to solve the problem. With problems such as these it is best to split the velocity into a vertical and horizontal component. To do this we construct a triangle, this can then be used with some trigonometry to find that the vertical component of the velocity is 10sin(50) ms^-1 and the horizontal component is 10cos(50) ms^-1. Now we must decide which of these components is important to the questions or whether we have to go through multiple steps to get to the answer. First consider the vertical direction. Write down all the known variables at the time the ball will hit the floor. This gives s(displacement) = 0 m as the ball has returned to its original position relative to the vertical direction, u(initial velocity) = 10sin(50) ms^-1 as calculated before, v(final velocity) = ? this is unknown, a(acceleration) = -9.81 ms^-2 this is due to gravity and the acceleration is in the opossite direction to the intial velocity and so is negative, t(time) = ? this is the unknown we are trying to calculate. From this information we need to consider the suvat equations of motion and decide whether we have enough information to use them. We have 3 pieces of information and an unknown so yes we do, the equations containing, s, u, a and t is s=ut+1/2at^2. We should rearrange the equation to find t, we know that s = 0 m so should substitute that in leaving, 0=ut+1/2at^2. We can then subtract ut from both sides to get, -ut=1/2at^2. Now divide both sides by t to get, -u=1/2at. Finally multiply both sides by 2 and divide both by a to get -2u/a=t. To finish we just have to substitute in our numbers. t = -(2 x 10sin(50)/-9.81 = 1.56 s