Question

1. a stunt car drives off a 150 m cliff at a speed of 45 mph. what is the cars velocity in the y - direction after 2 seconds?

29.43 m/s
-29.43 m/s
19.62 m/s
-19.62 m/s

Explanation:

Step1: Identify the formula

Use the kinematic equation \(v = v_0+at\). In the \(y -\)direction, the initial vertical velocity \(v_{0y}=0\ m/s\), and the acceleration \(a=-g=- 9.81\ m/s^{2}\) (negative because it is in the downward direction).

Step2: Substitute values into the formula

We want to find the velocity \(v_y\) at \(t = 2\ s\). Substitute \(v_{0y}=0\ m/s\), \(a=-9.81\ m/s^{2}\), and \(t = 2\ s\) into \(v_y=v_{0y}+at\).

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Answer:

\(-19.62\ m/s\)