a) The ball takes 4.29 s to hit the ground
b) The velocity at the impact is 65.3 m/s at [tex]40^{\circ}[/tex] below the horizontal
Explanation:
a)
The motion of the cannonball is a projectile motion, so it consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
We start by considering the vertical motion, to find the time of flight of the ball. The equation to use is the following:
[tex]s=u_y t+\frac{1}{2}at^2[/tex]
where, choosing downward as positive direction, we have:
s = 90 m is the vertical displacement (the height of the cliff)
[tex]u_y=0[/tex] is the initial vertical velocity (because the ball is fired horizontally)
t is the time of the fall
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
And solving for t, we find
:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(90)}{9.8}}=4.29 s[/tex]
So the ball takes 4.29 s to hit the ground.
b)
The velocity of the cannon ball has two components:
- The horizontal component of the velocity is constant during the entire motion (since there is no acceleration along the x-axis), and it is equal to the initial velocity at which the ball is fired:
[tex]v_x = 50 m/s[/tex]
- The vertical component of the velocity instead is given by
[tex]v_y = u_y + gt[/tex]
where
[tex]u_y=0[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
t is the time
The direction of this component is downward. Substituting t = 4.29 s, we find the vertical velocity at the impact:
[tex]v_y = 0 + (9.8)(4.29)=42.0 m/s[/tex]
And therefore, the velocity of the cannonball as it strikes the ground is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{50^2+(42.0)^2}=65.3 m/s[/tex]
And the direction is given by
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{42.0}{50.0})=40^{\circ}[/tex] below the horizontal (because the vertical velocity points downward)
Learn more about projectile motion:
brainly.com/question/8751410
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