Answer:
initial velocity = 29.4m/s
maximum height = 44.1m
Step-by-step explanation:
This problem is a projectile motion. The ball is tossed vertically from the ground, which means the angle between the ball and the horizontal is 90°
[tex]Time of flight (T) = \frac{2Usin\alpha }{g}[/tex]
[tex]\alpha =90[/tex]° (the ball is tossed vertically)
2Usinα = g*T
2U =gT/sinα
2U = 9.8* 6/ sin90°
2U=58.8
U = 58.8/2 =29.4
U=29.4m/s
maximum height attained (H) = [tex]\frac{U^{2}sin^{2}\alpha }{2g}[/tex]
=[tex]\frac{29.4^{2}*sin90^{2} }{2*9.8}[/tex]
maximum height =44.1m
