How to Calculate the Minimum Speed Required for a Stunt Car Jump

What is the minimum speed a stunt driver needs to make his car jump over 10 cars parked below a horizontal ramp?

The stunt driver must drive off the ramp with a minimum speed of 53.93 m/s to successfully clear the 10 parked cars.

Calculation Explanation

The trick driver needs to get an upward level free from 2.0 m and a flat distance of 24 m. To find the base speed required, we can involve the kinematic condition for vertical movement affected by gravity:
vf^2 = vi^2 + 2 * g * Δy
where vf is the last speed (extent of speed), vi is the underlying speed (size of speed), g is the speed increase because of gravity (9.8 m/s^2), and Δy is the adjustment of vertical level. We need the underlying speed (vi) so we can rework the condition:
vi = √(vf^2 - 2 * g * Δy)
For this situation, Δy = 2.0 m, g = 9.8 m/s^2, and vf = 0 (last speed is 0 in light of the fact that the vehicle lands on the ground).
vi = √(0 - 2 * 9.8 * 2.0) = √(- 39.2) = 6.27 m/s
This is the base vertical speed the vehicle should need to clear the 10 left vehicles. To find the base speed, we should likewise consider the even speed expected to cover the distance of 24 m.
v = √(vi^2 + vh^2)
where vh is the even speed. Since the even distance is 24 m and the hour of flight can be determined utilizing the upward movement condition, we can utilize the condition:
vh = d/t
where d is the flat distance (24 m) and t is the hour of flight.
t = √((2 * Δy)/g) = √((2 * 2.0)/9.8) = 0.447 s
vh = 24 m/0.447 s = 53.67 m/s
At long last, the base speed can be found as:
v = √(vi^2 + vh^2) = √(6.27^2 + 53.67^2) = 53.93 m/s (adjusted to 2 critical digits)
In this way, the trick driver should drive off the slope with a base speed of 53.93 m/s to clear the 10 left vehicles effectively.