A stunt driver wants to make a personal record breaking jump, launching her car

Question

A stunt driver wants to make a personal record breaking jump, launching her car over a line of 30 identical cars, each 5 ft tall. These cars are placed side by side with a total distance of 40 m. Her manager wants to ensure the jump is as safe and as dramatic as possible. The driver uses a horizontal ramp 6.00 m above the ground. The end of the ramp is directly above the first car in line.

(a) What is the stunt driver’s minimum speed when she leaves the ramp to make the jump?

(b) If her maximum speed off the ramp is 35 m/s, could she make a successful jump based on your answer to (a)? If not, please redesign the ramp to make it form an angle to the ground. Find the angle that would ensure a successful jump.

abe peena rd br er ar over a ie of 30 identical cars, each 5 ft tall. These cars are placed side by side with a total distance of 40 m. Her manager wants to ensure the jump is as safe and as dramatic as possible. The driver uses a horizontal ramp 6.00 m above the ground. The end of the ramp is directly above the first car in line (a) What is the stunt driver's minimum speed when she leaves the ramp to make the jump? (b) If her maximum speed off the ramp is 35 m/s, could she make a successful jump based on your answer to (a)? If not, please redesign the ramp to make it form an angle to the ground. Find the angle that would ensure a successful jump

Explanation / Answer

Given,

D = 40 m ; H = 6 m ;

We know from eqn of motion

S = ut + 1/2 at^2

when S = H ; u = 0 ; a = g

t = sqrt (2 H/g) = sqrt (2 x 6/9.81) = 1.11 s

we know that,

D = Vx t => Vx = D/t

Vx = 40/1.11 = 36.04 m/s

Hence, Vx = 36.04 m/s

b)No she will not be able in making a successful jump off the ramp as will smash into any car midway since she will not be able in covering a distance of D = 40 m with the speed of 35 m/s.

still the horizontal distance is D = 40 m

D = V cos(theta) t

t = V sin(theta)/g

D = V^2 cos(thets) sin(thets)/g

sin(2theta) = D g/V^2 = 40 x 9.81/35^2 = 0.320

2theta = sin^-1(0.320) = 18.68 deg

theta = 18.68/2 = 9.34 deg

Hence, theta = 9.34 deg

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