# Thread:

1. ## Projectile Motion Question

Alright, so I have the following projectile motion problem:

A daredevil jumps a canyon 12 m wide, to do so, he drives a car up a 15 degree incline. What minimum speed must he achieve to clear the canyon, and if he jumps at this minimum speed, what will his speed be when he reaches the other side?

I actually have the answer to the first question, I got it to be 15.46 using the two projectile launched at an angle formulas, subbing in 12.42/vi for t in the horizontal formula. What I'm looking for is how to solve the second part of the problem. Thanks for the help =)

2. The speed will be exactly the same as the starting speed. An easy way to see this is by looking at the kinetic energy of the car: as the car rises, its kinetic energy is converted to gravitational potential energy, which is changed back into kinetic energy as the car goes down again. Since the car rises and falls the same distance, the same amount of energy is converted so the car ends up at the speed it started at.

3. You could try using to formula $v^{2}_{x}=v^{2}_{0_{x}}+2a_{x}\Delta x$

where $v^{2}_{x}$ is the final horizontal velocity, $x^{2}_{0_{x}}$ is the initial horizontal velocity (plug in what you found for the first part here), $a_x$ is the horizontal acceleration (im assuming this is zero?), and $\Delta x$ is the horizontal displacement, or 12m in this case.

4. You're right, there is no horizontal acceleration since there are no horizontal forces mentioned.

$v_{0x}^2 = v_{fx}^2 + 2a_x \Delta x$

$v_{0x}^2 = v_{fx}^2 + 2(0)\Delta x$

$v_{0x}^2 = v_{fx}^2$

$v_{0x} = v_{fx}$

But this is only in the x direction, the y-direction's change in velocity DOES affect the final velocity.

5. Originally Posted by Crysolice
Alright, so I have the following projectile motion problem:

A daredevil jumps a canyon 12 m wide, to do so, he drives a car up a 15 degree incline. What minimum speed must he achieve to clear the canyon, and if he jumps at this minimum speed, what will his speed be when he reaches the other side?

I actually have the answer to the first question, I got it to be 15.46 using the two projectile launched at an angle formulas, subbing in 12.42/vi for t in the horizontal formula. What I'm looking for is how to solve the second part of the problem. Thanks for the help =)
v sin15=y
v cos15= x

x y
delta x=1/2at^2+vt delta y= 1/2at^t + vt
delta x=vt 0=1/2(-9.8)t^2=vt
12=v*cos15*t 0=-4.9t^2+v sin15 t
12.37/v=t 0=-4.9(12.37/v)^2 + v sin15 (12.37/v)
749.8/v^2 = v sin15 (12.37/v)
749.8/v^2 = 3.20v/v
749.8 = 3.20v^2
239.3125=v^2
15.46
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