A projectile of mass 0.257 kg is shot from
a cannon, at height 6.4 m, as shown in the
figure, with an initial velocity vi having a
horizontal component of 6.4 m/s.
The projectile rises to a maximum height of
y above the end of the cannon’s barrel and
strikes the ground a horizontal distance x
past the end of the cannon’s barrel.
Find the magnitude of the velocity vector
when the projectile hits the ground. The
acceleration of gravity is 9.8 m/s2 .
Answer in units of m/s

Question

A projectile of mass 0.257 kg is shot from
a cannon, at height 6.4 m, as shown in the
figure, with an initial velocity vi having a
horizontal component of 6.4 m/s.
The projectile rises to a maximum height of
y above the end of the cannon’s barrel and
strikes the ground a horizontal distance x
past the end of the cannon’s barrel.
Find the magnitude of the velocity vector
when the projectile hits the ground. The
acceleration of gravity is 9.8 m/s2 .
Answer in units of m/s

john_es · Answer

I think some more data is needed. Could you post the graph that problem cites?

anonymous · Answer

In the picture it shows the cannon at a 44 degree angle

anonymous · Answer

Hmmm well I tried that and the answer was incorrect is there another way to do it?

john_es · Answer

Well, we can try with kinematics.

anonymous · Answer

Well it makes sense but the answers are still wrong :/

john_es · Answer

Rewritting.

john_es · Answer

From before,
$$v_{0x}=v_0\cosθ⇒v_0=v_{0x}\cosθ=8.897m/s\
v_{0y}=v_0\sinθ=6.18m/s$$
Now,
$$y=y_0+v_{0y}t−\frac{1}{2}gt^2⇒0=6.4+6.18t−4.9t^2⇒t=1.936 s\
v_y=v_{0y}−gt=−12.79 m/s$$ (The same result obtained with energy but it was the y component, this was my mistake, to consider it was the tota velocity)
$$v=\sqrt{v^2_{0x}+v^2_y}=14.3\ m/s$$

john_es · Answer

Now, it is ok.

anonymous · Answer

Thank you :)