a) α=53.13°
b) V=15 m/s a=(-9.8 m/s^2)
c) Height= 15.9 m Velocity= 17.7 m/s
Explanation:
I'm going to be including a picture of my working and explaining it here since I find it to be easier to teach.
In physics, in order to solve a problem, a drawing must be made for each question in order to comprehend what it asks and picture it better when solving equations, which is what I did and I recommend you do it as well when solving questions in the future, as well as setting a frame of reference for all motion going on, you'll soon see why.
a) In order to understand projectile motion, one must understand that motion in two dimensions, and therefore the forces that cause it, have two components: one for each axis of movement. I have chosen that the x-axis will determine horizontal movement, and the y-axis will determine vertical movement. Now, we have a velocity with a magnitude of 25 m/s that goes diagonally into the air at an unknown angle with unknown vertical and horizontal magnitudes. These characteristics sound like a right-angled triangle, and like such, we can draw one where its hypotenuse equals the initial velocity's magnitude, its length its horizontal magnitude (labelled Vx), its height its vertical magnitude (labelled Vy) and α the angle between the triangle's hypotenuse and length (drawn in the first picture to the right). Now, since velocity is now represented by a right-angle triangle, we can also use Pythagoras' theorem and SohCahToa on it.
Through SohCahToa we can now find an expression for the velocity's horizontal magnitude, this being Vx/V= cos (α), where we have two unknown variables: Vx and α. Observing the system we just drew, we can see that the only outside force acting on the water after leaving the hose is gravity, which pulls the water downwards and has no horizontal components, this means that the horizontal movement in our system will remain constant unless its physically stopped, as said by Newton's first law (When undisturbed a force in motion will stay in motion). Therefore the horizontal velocity of the water leaving the hose will be equal to the horizontal velocity of the water when it reaches the building, of which we know the distance from the hose and time of arrival. With this knowledge we can make a simple velocity calculation: V=distance/time, and find that the horizontal velocity of the water is always 15 m/s. Now we replace that value in the original horizontal velocity expression Vx/V= cos (α) and find α since we know both V (25 m/s) and Vx (15 m/s)
b) No calculations are really necessary in this question, since the highest point of the trajectory means that all upward motion has now stopped, or in other words: there is no more upwards velocity (Vy=0). Therefore, if one were to carry out Pythagoras' theorem at this point in the trajectory (hypotenuse equals the square root of the other two sides each squared), it would end up being equal to the horizontal velocity (15 m/s).
The only acceleration on this system is that of gravity, since its the only force acting on the water while in the air, and therefore its -9.8 m/s (or 9.81 or 10 depending on what values your teacher asks you to use) which is negative since it pulls the water downward according to my frame of reference.
c) Now, in order to determine final velocity and height we need to find the vertical magnitude of the initial velocity, which we can easily do with the triangle we used in a), this being Vy/V = sen (α) which we can change to Vy = V x sen (α). Replacing V and α we find that Vy= 20 m/s. Now, in order to find the height we can now turn to uniformly accelerated motion (UAM), since if we were to look at a drawing of Vy's movement on its own with gravity, it would travel in a straight line upwards and then fall downwards. Using UAM's formula for distance: y(t)= initial y value + initial velocity (Vi) x t + 1/2 x acceleration x t squared, and replacing it with the values we already have at the time the water reaches the building, the equation ends up looking like this: y(3s) = 0m (ground) + 20 m/s x 3s + 1/2 x (-9.8 m/s^2) x 3s^2. The result of this is y(3s)= 15.9 m.
The final vertical velocity (Vyf) can be found with another equation from UAM, which is Vf(t) = Vi + a x t. Replacing values we end up with Vf(3s) = 20 m/s + (-9.8 m/s^2) x 3s. The result is Vf(3s) = -9.4 m/s. Using Pythagoras with the values from when the water reached the building we get the final velocity, which is 17.7 m/s.
