Question

A car drives straight off the edge of a cliff that is h =42 m high. The police at the scene of the accident note that the point of impact is d =72 m from the base of the cliff. Here the initial speed v0 is unknown. Ignore air resistance. The magnitude of the gravitational acceleration is 9.8 m/s2. Choose the RIGHT as positive x-direction. Choose UPWARD as postitive y-direction Keep 2 decimal places in all answers. How long did it take to hit the ground?

Answer 1

Answer:

It took 2.93 s$v_{f,y} = v_{0,y} + a*t \\-28.69 = 0 - 9.8 * t \\t = 2.93 s$

Explanation:

the speed has 2 components in x and in y

v0,y = 0

v0,x = unknown

distance in y = 42 m

distance in x = 72 m

Y:

The car fell 42 m with an initial speed of 0 m/s, gravitational acceleration 9.8 m/s^2

With that information you can calculate the final speed at the moment of impact

$v_{f,y}^{2} = v_{0,y}^2 + 2g*h \\v_{f,y}^{2} = 2*9.8*42 \\v_{f,y}^{2} = 823.2 \\v_{f,y} = \sqrt{823.2} \\v_{f,y} = -28.69 m/s$

I put a minus because it says upward is positive and the speed direction is downward.