30 POINTS, BRAINLIEST, PLEASE HAALLPPP TT^TT PLEASE don`t respond without an answer!
A basketball is thrown upward from a height of 1.8 m at a velocity of 12 m/s. How long until it reaches the ground?
(Recall: h(t)=-4.9t^2-v_0 t+h_0, where v_0 is initial velocity and h_0 is initial height.)

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

With these questions it's best to just memorise the suvat equations and either draw or imagine the actions involved, that was you can tell what piece of given information translates to which variable. For example; I know that I am looking for max height, so when a ball is at its highest point, it can't be moving up anymore (thus V = 0). I also know that it is moving upwards against gravity, so gravity will be decelerating the ball (therefore a=-g). In a paper, it may as for assumptions with this question, a good answer would be no air resistance or movement due to any other external forces. I hope I helped :)

The volume is 56.8 triangular the

8-2x+3x2=6 and six is your answer