2 dimensional projectile motion

[christy77]

Homework Statement

A stunt driver wants to make his car jump over 10 cars parked side by side below a horizontal ramp. if the horizontal distance is 80 ft and vertical distance is 6 ft., what is the minimum speed must the driver drive off the ramp?

Homework Equations

vx=v°*cos(θ)
v°y=v°sinθ

The Attempt at a Solution

80/cos45=113
6/sin45=8.5

 

[TheKracken]

Are you going to attempt this problem?

 

[christy77]

i did

 

[nejibanana]

You should look up Kinematic Equations in your book or with google.

 

[Nickie]

Using the equations for horizontal and vertical velocity, we can calculate the minimum speed the driver must drive off the ramp to successfully clear the 10 parked cars. Based on the given horizontal distance of 80 ft and vertical distance of 6 ft, we can assume the ramp is at an angle of 45 degrees. Using the formula vx = v° * cos(θ), where vx is the horizontal velocity, v° is the initial velocity, and θ is the angle of the ramp, we can calculate the minimum horizontal velocity needed to travel 80 ft. Similarly, using the formula v°y = v° * sin(θ), where v°y is the vertical velocity, we can calculate the minimum vertical velocity needed to travel 6 ft. The minimum speed required to successfully clear the 10 parked cars would be the total velocity, which can be calculated using the Pythagorean theorem (v = √(vx² + v°y²)). Therefore, the minimum speed the driver must drive off the ramp is approximately 113 ft/s.