10. A cannon elevated at an angle of 35º to the
horizontal fires a cannonball, which travels the path
shown in the diagram below. [Neglect air resistance and
assume the ball lands at the same height above the
ground from which it was launched.]
If the ball lands 7.0 × 10² meters from the cannon 10.
seconds after it was fired, what is the horizontal
component of its initial velocity?
(A) 7.0 m/s (B) 35 m/s (C) 49 m/s (D) 70. m/s
8 answers
The horizontal component of velocity remains constant, so divide 700 m by 10 s.
49 m/s
49 m\s
49 m/s my man
Bro, i put 49 m/s, but it's actually 70 m/s
its 70
70 m/s
the initial HORIZONTAl speed is 70 m/s
and the initial VERTICAL speed is 49 m/s
ignoring air resistance we know that the horizontal velocity is constant so you can use the formula d/t=v
700/10= 70 m/s
to find the initial vertical speed you can plug the 70m/s into the tangent function which is tan θ = vi(y-axis) / vi(x-axis)
tan 35° = vi(y-axis) / 70
rearrange it to solve for vi (y-axis)
vi(y-axis) = 70 * tan 35°
vi (y-axis) = 49 m/s
** keep in mind y axis is for the vertical
**and x axis is for the horizontal
and the initial VERTICAL speed is 49 m/s
ignoring air resistance we know that the horizontal velocity is constant so you can use the formula d/t=v
700/10= 70 m/s
to find the initial vertical speed you can plug the 70m/s into the tangent function which is tan θ = vi(y-axis) / vi(x-axis)
tan 35° = vi(y-axis) / 70
rearrange it to solve for vi (y-axis)
vi(y-axis) = 70 * tan 35°
vi (y-axis) = 49 m/s
** keep in mind y axis is for the vertical
**and x axis is for the horizontal
0 answers