Question

A cannon ball was fired from the top of a 100 m tall cliff. The cannon was aimed at 25 degrees below the horizontal and the ball left the gun with a speed of 3x10^2 m/s. Find the impact velocity of the cannon ball.

Answer 1

to horizontal direction \(V_x = V cos 25\) \(x_0 =0 \) \(a_x =0\)
to vertical direction \(V_y = Vsin25\) \(y_0= 0\) y = 100
but I don't understand the sentence " find the impact velocity of the cannonball" do they ask you find just \(V_x ~~and~~V_y\)

Answer 2

I don't know.

Answer 3

ok, do as we think, to horizontal direction \(V _x= Vcos 25= 3*10^2 cos 25 = 271.89m/s\) that 's it
to vertical direction \(V_{final}^2= V sin25 + as= 3*10^2sin25 + (9.8*100)= 1106.79m/s \)

Answer 4

so \(V_y = 33.26m/s\)

Answer 5

that's all I can do for you. I am not sure about the logic.

Answer 6

But isn't it -25?