a ball is thrown straight up with an initial speed of 30m/s. how much tim dose it take o reach the top of its trajectory?
Use the equation Vy = Viy + a(T)
Viy = 30m/s
a = -9.82m/(s^2)
At the top of the ball’s trajectory, assume that its velocity will be zero.
Vy = 0
Rearrange, plug in the variables into your equation and solve
Vy = Viy + a(T) (Vy - Viy)/(a) = T (0-30m/s)/(-9.82) = T = 3.055s