Cannonball trajectory?

Question

A cannonball is shot from a hill 10 meters higher than the surrounding ground.

a. If the firing trajectory is 0 degrees from the horizontal, and the firing velocity 50 m/sec, how far will the cannon ball travel?
b. If the cannon’s trajectory is 30 degrees instead, how far will the cannon ball travel now?
c. If the cannon’s trajectory is 45 degrees how far will its range be?
d. If the cannon is moved to level ground and fired at 45 degrees, how far will its range be?

I don't know what equations should be used for this problem and have no idea where to start. Any help would be great.

Answer 1

X(t) = Xo + vt + 1/2at^2
and Y (t) = y0 + vt - 1/2gt^2

a) at zero degrees, the ball falls 10 meters before it hits ground. Thus using y equation, Solve for time.

-10 = 0 + 0 -0.5 (9.8) t^2

Plug that T back into X equation to get distance.
X = 0 + 50t + 0.

b. At 30 or 45 degrees, you want to find the X and Y components of the velocity. It would be sin30 or 45 times 50 to get Y velocity component.
50cos 30 or 45 to get X velocity component. After you get these, make the neccesary corrections in the X and Y equation above. For example of 30,
Y (t) = y0 + vt - 1/2gt^2
-10 = 0 + 50sin30t - 1/2gt^2
Solve for t, plug into X
X(t) = Xo + vt + 1/2at^2
X(t) = 0 + 50cos30 T + 0.

for D, the displacement is 0 so its no longer -10.

Note: the 2 equations only work for constant acceleration.

Answer 2

x=(-b±SQRT(b^2*4ac))/2ab=ua=ax=tc=s

the +- has to be varied for a positive answer for b it is negative and for c it is positive, or something like that

if you know s=ut+1/2at2 then you can do the above equation to work out the time it takes.
in this case a = -9.8
c = either -10 or 0 for d
b = COS (for b 30 or for c 45)*50 (the vertical component of the balls speed)

once you work out how long it takes to reach the ground you times it by SIN(for b 30 or for c 45)*50 (which is the horizontal component of its speed)


for a it is very simple as it is just 10/9.8*50
for d do the same as above but as said the end distance is 0.

A)51.02040816
B)3962.346
C)10503.89639
D)558.7479148

i hope the answers are right.

Answer 3

Use:

s=ut+1/2at^2

and solve horizontally and vertically.

s = displacement from point of origin (so horizontally this would be the unknown you are trying to find out and vertically it would be 10 metres)
u = initial velocity
t = time
a = acceleration

Have fun!