Question

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Answer 1

\[h(t) = -4.9t^2 + v _{0}t + h _{0}\]

v0 = initial velocity = 0
h0 = initial height = 150
h(t) = height after time (t) = 0

It is 0 since we are looking for when it hits the ground, which is 0.

\[0 = -4.9t^2 + (0)t + 150\]
\[0 = -4.9t^2 + 150\]
\[4.9t^2 = 150\]
\[t^2 = \frac{ 150 }{ 4.9 }\]
\[t = \sqrt (\frac{ 150 }{ 4.9 })\]
t is in seconds

Answer 2

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Answer 3

Okay good