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A car drives horizontally off a 63-m-high cliff at a speed of 29 m/s . Ignore air resistance.

1. How long will it take the car to hit the ground? (in seconds)
a) 2.9
b) 4.0
c) 3.6
d) 1.9
e) 5.2

2. How far from the base of the cliff will the car hit? (in meters)
a) 54
b) 190
c) 100
d) 210
e) 250

I already know the answers to these questions, but if I could have an explanation as to how to solve these problems I would greatly appreciate it!!

Respuesta :

LammettHash

The car's vertical position [tex]y[/tex] at time [tex]t[/tex] is

[tex]y = 63\,\mathrm m - \dfrac12 gt^2[/tex]

since it starts 63 m above the ground, and after leaving the cliff it accelerates downward due to gravity.

Its horizontal position [tex]x[/tex] is

[tex]x = \left(29\dfrac{\rm m}{\rm s}\right) t[/tex]

since the car leaves the cliff horizontally at 29 m/s, and is not influenced by any other acceleration in this plane.

1. Solve for [tex]t[/tex] such that [tex]y=0[/tex].

[tex]63\,\mathrm m - \dfrac12 gt^2 = 0 \implies t = \sqrt{\dfrac{126\,\rm m}g} \approx \boxed{3.6}\,\rm s[/tex]

2. Solve for [tex]x[/tex] at this value of [tex]t[/tex].

[tex]x = \left(29\dfrac{\rm m}{\rm s}\right) \sqrt{\dfrac{126\,\rm m}g} \approx 104\,\rm m \approx \boxed{100}\,\rm m[/tex]

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