Respuesta :
Answer:
Final velocity = 68.78 m/s
Angle of impact = 274.17°
The ball lands = 35 m from the cliff
Explanation:
Given:
[tex]initial\ x-velocity\ (v_o_x)=5\ m/s[/tex]
[tex]time\ (t)=7\ s[/tex]
[tex]acceleration=gravity\ (g)=-9.8\ m/s^2[/tex]
First we break the motion into horizontal component (x-component) and vertical component (y-component).
Horizontal component:
Since there is no additional horizontal force, based on the first Newton Law, the velocity ([tex]v_o_x[/tex]) will remain constant (linear motion with constant velocity).
[tex]\boxed{final\ x-velocity\ (v_x)=v_o_x}[/tex]
[tex]\bf v_x=5\ m/s[/tex]
[tex]\boxed{horizontal\ distance\ (x)=x_o+v_x\cdot t}[/tex]
[tex]x=0+5(7)[/tex]
[tex]\bf x=35\ m[/tex]
Vertical component:
Gravitational force will increase the velocity ([tex]v_o_y[/tex]) with a constant rate (linear motion with constant acceleration).
[tex]\boxed{final\ y-velocity\ (v_y)=v_o_y+at}[/tex]
[tex]v_y=0+(-9.8)(7)[/tex]
[tex]\bf v_y=-68.6\ m/s[/tex]
Now, we combine the final horizontal and vertical component to find the magnitude and angle of the final velocity.
[tex]\boxed{final\ velocity\ (v)=\sqrt{v_x^2+v_y^2} }[/tex]
[tex]v=\sqrt{(5)^2+(-68.6)^2}[/tex]
[tex]\bf v=68.78\ m/s[/tex]
[tex]\boxed{tan\theta=\frac{v_y}{v_x} }[/tex]
[tex]tan\theta=\frac{-68.6}{5}[/tex]
[tex]tan\theta=-13.72[/tex]
(Since y is negative and x is positive, therefore θ is in Quadrant 4)
[tex]\bf\theta=274.17^o[/tex]
Answer: 240.1m
Explanation:
Here is the distance formula.
[tex]d=\frac{1}{2} gt^{2}[/tex]
Then plug in your numbers. (9.8 is gravity)
[tex]d=\frac{1}{2} (9.8m/s)(7s)^{2}[/tex]
d= 240.1m
