Question #4823

A daredevil jumps a canyon 12 m wide. To do

so, he drives a car up a 10 incline.

What minimum speed must he achieve to

clear the canyon?

The acceleration of gravity is 9.81 m/s2 .

Answer in units of m/s

so, he drives a car up a 10 incline.

What minimum speed must he achieve to

clear the canyon?

The acceleration of gravity is 9.81 m/s2 .

Answer in units of m/s

Expert's answer

To find this, you must first find the peak altitude which occurs at

the

moment his vertical velocity is zero.

Vy(t) = Vyo -gt = 0 => t =

Vyo/g

We know the time from ground to peak altitude is equal to the

time

from peak altitude to ground. So the total time of flight is 2*t

=

2*Vyo/g.

The distance the car travels is Vxo*2*t = Vxo*2*Vyo/g = 12

meters.

Use tan10=Vyo/Vxo => Vyo = Vxo*tan10

Substitute into the

distance equation:

[2*Vxo^2*tan10]/g = 12

Vxo = sqrt[12g/2*tan10] =

sqrt[117.72 / 0.3527] = 18.27 m/s, but that

is just the horizontal

component

Vo = Vxo/cos10 = 18.55 m/s

the

moment his vertical velocity is zero.

Vy(t) = Vyo -gt = 0 => t =

Vyo/g

We know the time from ground to peak altitude is equal to the

time

from peak altitude to ground. So the total time of flight is 2*t

=

2*Vyo/g.

The distance the car travels is Vxo*2*t = Vxo*2*Vyo/g = 12

meters.

Use tan10=Vyo/Vxo => Vyo = Vxo*tan10

Substitute into the

distance equation:

[2*Vxo^2*tan10]/g = 12

Vxo = sqrt[12g/2*tan10] =

sqrt[117.72 / 0.3527] = 18.27 m/s, but that

is just the horizontal

component

Vo = Vxo/cos10 = 18.55 m/s

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