Answer:
ball clears the net
Explanation:
[tex]v_{o}[/tex] = initial speed of launch of the ball = 20 ms^{-1}
[tex]\theta[/tex] = angle of launch = 5 deg
Consider the motion of the ball along the horizontal direction
[tex]v_{ox}[/tex] = initial velocity = [tex]v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}[/tex]
[tex]t[/tex] = time of travel
[tex]X[/tex] = horizontal displacement of the ball to reach the net = 7 m
Since there is no acceleration along the horizontal direction, we have
[tex]X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}[/tex] Eq-1
Consider the motion of the ball along the vertical direction
[tex]v_{oy}[/tex] = initial velocity = [tex]v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}[/tex]
[tex]t[/tex] = time of travel
[tex]Y_{o}[/tex] = Initial position of the ball at the time of launch = 2 m
[tex]Y[/tex] = Final position of the ball at time "t"
[tex]a_{y}[/tex] = acceleration in down direction = - 9.8 ms⁻²
Along the vertical direction , position at any time is given as
[tex]Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\[/tex]
Since Y > 1 m
hence the ball clears the net
