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: Â
- A uniform motion (constant velocity) along the horizontal direction Â
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction Â
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, Â 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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