Question

10 lars is standing near the edge of a 90 - meter cliff. he throws a ball upward, but does not catch it, and it falls to the bottom of the cliff face. from when he threw the ball upward to when it hit the ground below, 5.55 seconds passed. what was the initial vertical velocity of lars throw? 43 m/s 11 m/s - 43 m/s - 11 m/s

Explanation:

Step1: Use the kinematic equation

The kinematic equation for vertical displacement is \(y = y_0+v_0t-\frac{1}{2}gt^2\). Here, \(y = 0\) (ground level), \(y_0=90\) m (height of the cliff), \(t = 5.55\) s, and \(g = 9.8\) m/s².
Substituting the values into the equation: \(0=90 + v_0\times5.55-\frac{1}{2}\times9.8\times(5.55)^2\).

Step2: Simplify the equation

First, calculate \(\frac{1}{2}\times9.8\times(5.55)^2\).
\(\frac{1}{2}\times9.8\times(5.55)^2=4.9\times30.8025 = 150.93225\).
The equation becomes \(0=90 + 5.55v_0-150.93225\).

Step3: Solve for \(v_0\)

Rearrange the equation: \(5.55v_0=150.93225 - 90\).
\(5.55v_0 = 60.93225\).
Then \(v_0=\frac{60.93225}{5.55}\approx11\) m/s.

Answer:

11 m/s