Could someone please help me with the following projectiles question?
Tam throws a ball across a field a horizontal distance of 100m. She throws it at an angle of 30 degrees. It lands at the same height (ignore air resistance). At what speed does the ball leave her hand?
The answer is in the spoiler below
Spoiler
The answer is 34 m/s
Thank you!
Let's break this down completely. I'm assuming that you recognise this is a constant acceleration in 2D problem (vertical acceleration =g downwards, horizontal acceleration = 0. This part is important; please be sure of this first)
What do we need for a horizontal distance? Well, because the acceleration is zero in that direction (the vertical direction doesn't affect the horizontal direction to a very good approximation; this has been verified by experiment and it follows from the fact that the gravitational force is vertical. I say approximation because the Earth is a rotating sphere and that complicates things but let's disregard that), the distance is just the horizontal velocity * time. The horizontal velocity is related to the speed and the angle, which you have.
So we have 100 = v(x)t = u*cos 30*t
The problem is, this is one equation in two unknowns, so we need another piece of information. This is where we use the vertical motion.
If the ball is thrown across a horizontal field, its total vertical displacement is identically zero. This is also important. In these questions, you need to recognise what words mean what mathematically. That's the art of problem solving: converting given information into useful information.
Anyway, you now have the vertical displacement and the acceleration. Remember how we still don't know the speed at the time? Let's write out another equation that uses them so that we can solve for them. The relevant constant acceleration equation that we want now is the one that involves the initial vertical velocity (to include the speed), the time and the other two known quantities, so our equation becomes 0 = v(y)t - 1/2 at^2.
t isn't zero because we're not interested in the instant the ball is thrown.
So v(y) = 1/2 at -> u sin 30 = 1/2 at, u = at
Plug this into the other equation and solve for the speed.
Of course, you can do this in general and with a bit of algebra, you can find that the general formula for the range for a projectile with no gain in height is R = v^2 sin (2*theta) / g. I'll leave that up to you. Hint: you'll need a double angle trig identity somewhere for this.