The time of flight of the ball is 3.08 s
Explanation:
The motion of the cannonball is a projectile motion, consisting of two motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration along the vertical direction
Here we can solve the problem just by analyzing the horizontal motion. In fact, we know that:
u = 24 m/s is the initial velocity of the ball
[tex]\theta=40^{\circ}[/tex] is the angle of projection
d = 56.6 m is the range of the ball (the horizontal distance covered)
First of all, we can find the initial velocity in the horizontal direction:
[tex]v_x = u cos \theta = (24)(cos 40)=18.4 m/s[/tex]
We know that this horizontal velocity is constant, so it is related to the range by:
[tex]d=v_x t[/tex]
where t is the time of flight of the ball; and solving for t, we find
[tex]t=\frac{d}{v_x}=\frac{56.6}{18.4}=3.08 s[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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