The ball reaches the top of its trajectory after 1.25 s
Explanation:
The motion of the ball is a projectile motion, which consists of two independent motions:
- A uniform motion (at constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
To solve the problem we are only interested in the vertical motion of the ball. The vertical velocity of the ball at time t is given by
[tex]v_y = u_y -gt[/tex]
where
[tex]u_y[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
t is the time
The ball reaches its maximum heigth when its vertical velocity is zero:
[tex]v_y = 0[/tex]
Also, the initial vertical velocity is given by
[tex]u_y = u sin \theta[/tex]
where
u = 18 m/s is the initial speed
[tex]\theta=43^{\circ}[/tex] is the angle of projection
Solving the equation for t, we find the time at which the ball reaches the maximum height:
[tex]0=u sin \theta -gt\\t=\frac{u sin \theta}{g}=\frac{(18)(sin 43^{\circ})}{9.8}=1.25 s[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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