A grenade is thrown up into the air with an initial velocity of 27 m/s. How long does it take to return to the thrower's hand?

I got 5.4 s, but I am very unsure.

1 answer

well, you know that

y(t) = 27t - 4.9t^2
y=0 when

27t-4.9t^2 = 0
t = 27/4.9 = 5.51

You are close, but I wonder how you got what you did...

Ah. I see you used g = 10 m/s^2, not 9.8

OK, you are correct.