Question

A stunt man drives a car at a speed of 20 m/s off a 26-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20∘

Part A:

How far from the base of the cliff does the car land?

Express your answer to two significant figures and include the appropriate units.

Part B:

What is the car's impact speed?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Initial Velocity =20m/s at 20 degrees

so $u_{x}=ucos20=20cos20 =18.79m/s$

$u_{y}=usin20=20sin20 =6.84m/s$

The car will rise a little then falls

Consider the vertical motion of car

Rise

Use Formula $v^{2}-u^{2}=2as$

$v_{y}^{2}-u_{y}^{2}=-2g*y$

$(0m/s)^{2}-(6.84m/s)^{2}=-2*9.81m/s^{2}*y$ (at max height velocity becomes zero)

$0^{2}-6.84^{2}=-2*9.81*y$

$6.84^{2}/19.62=y$

${\color{Red} y=2.385m}$

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Time taken to rise

$v=u+at$

$v_{y}=u_{y}-gt_{r}$

$0m/s=6.84m/s-9.81m/s^{2}*t_{r}$

$0=6.84-9.81t_{r}$

${\color{Red} t_{r}=0.7s}$

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Falling

Falling Height =y +cliff height =2.385m +26m =28.385m

Time for falling

Use formula $s=ut+1/2at^{2}$

$H=u'_{y}t_{f}+1/2gt_{f}^{2}$

Initial velocity is Zero

$H=1/2gt_{f}^{2}$

$28.385m=0.5*9.81m/s^{2}*t_{f}^{2}$

$28.385=0.5*9.81*t_{f}^{2}$

${\color{Red} t_{f}=2.41s}$

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Total time,$t=t_{r}+t_{f}=0.7s+2.41s=3.11s$

${\color{Red} t=3.11s}$

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Consider the horizontal motion of car

There is no acceleration in horizontal direction

Speed =Distance /time

Distance=Speed*time

$x=u_{x}*t$

$x=18.79m/s*3.11s$

Part A: ANSWER: ${\color{Red} x=58.44m}$

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Part B

$u_{x}$ remains same

so $u_{x}=v_{x}=18.79m/s$

Vertical motion

Use Formula $v^{2}-u^{2}=2as$

$v_{y}^{2}-u'_{y}^{2}=2gH$

$v_{y}^{2}=2gH$

$v_{y}=\sqrt{2gH}$

$v_{y}=\sqrt{2*9.81m/s^{2}*28.385m}$

$v_{y}=23.6m/s$

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$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{(18.79m/s)^{2}+(23.6m/s)^{2}}$

ANSWER: ${\color{Red} v=30.17m/s}$

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