Answer:
The height reached by the ball=0.459 m
Explanation:
We are given that
[tex]\theta=30^{\circ}[/tex]
Initial speed, u=6 m/s
We have to find the maximum height reached by ball.
Maximum height reached by ball
[tex]h=\frac{u^2sin^2\theta}{2g}[/tex]
Where [tex]g=9.8m/s^2[/tex]
Substitute the values
[tex]h=\frac{6^2sin^230^{\circ}}{2\times 9.8}[/tex]
[tex]h=\frac{36\times \frac{1}{4}}{2\times 9.8}[/tex]
[tex]h=0.459 m[/tex]
Hence, the height reached by the ball=0.459 m
