A ball is thrown upward from a height of $9 \mathrm{~m}$ at an initial speed of $57 \mathrm{~m} / \mathrm{sec}$. Acceleration resulting from gravity is $-9.8 \mathrm{~m} / \mathrm{sec}^{2}$. Neglecting air resistance, solve for the velocity $v(t)$ and the height $h(t)$ of the ball $t$ seconds after is is thrown.
Velocity $v(t)=$ meters/second
Height $h(t)=$ meters
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Answer
\(\boxed{h(t) = -4.9t² + 57t + 9}\) m is the height of the ball at any time t.
Step 1 ：Given that the initial velocity (u) is 57 m/s, the acceleration due to gravity (a) is -9.8 m/s², and the initial height (h₀) is 9 m.
Step 2 ：We can calculate the velocity (v) at any time (t) using the equation \(v(t) = u + at\). Substituting the given values, we get \(v(t) = 57 - 9.8t\) m/s.
Step 3 ：We can calculate the height (h) at any time (t) using the equation \(h(t) = ut + 0.5at² + h₀\). Substituting the given values, we get \(h(t) = -4.9t² + 57t + 9\) m.
Step 4 ：\(\boxed{v(t) = 57 - 9.8t}\) m/s is the velocity of the ball at any time t.
Step 5 ：\(\boxed{h(t) = -4.9t² + 57t + 9}\) m is the height of the ball at any time t.
