Asked by Diane
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.
I got 5.4 s, but I am very unsure.
Answers
Answered by
Steve
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.
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.