projectile range equation with spring force

Question

A group of bored kids at a skate park decide to build a launcher out of elastic exercise bands (i.e., stretchy strips of rubber or a bungee cords) and a wood ramp. They tie several bands to two closely spaced trees on a flat field and place the ramp in between the trees. The 50-kg "rider" stands on their skateboard at the bottom of the ramp as the other kids stretch the elastic bands back, hook them around the back of rider and then release the bands so that they snaps back, launching the rider up the ramp. Approximate the bands as acting like a spring* with a spring constant of k = 1000 N/m that has been stretched by 2.0 meters. The ramp is angled at 30º above the horizontal and is 3.0 meters long. Treat the skateboard rolling on the ramp as frictionless and neglect air resistance. Calculate the horizontal distance between the back end of the ramp (the end where the rider goes airborne) and the location that the skateboard lands on the ground.

Answer 1

θ = 30°, x = 3m

Elastic Potential Energy = Initial Kinetic Energy (KE₀)

½kx² = ½(1000 N/m)(2 m)² = 2000 J = KE₀

Gravitational Potential Energy at Launch

PE_g = mgh where h = x sin(θ) = 3 sin(30°) = 1.5 m

So PE_g = (50 kg)(9.8 m/s²)(1.5 m) = 735 J

Going up the ramp, the kinetic energy of the skateboarder

decreases by this amount.

Kinetic Energy at Launch (KE_L) Determines Launch Velocity (v)

KE_L = KE₀ – PE_g = 2000 – 735 = 1265 J = ½mv²

v = √(2KE_L/m) = √(2(1265)/50) = 7.1134 m/s

Projectile Motion from the end of the ramp

Components of velocity v

vₓ = v cos θ = 7.1134 cos(30) = 6.1604 m/s

vᵧ = v sin θ = 7.1134 sin(30) = 3.5567 m/s

Time of flight (t)

y – y₀ = vᵧt – ½gt² which means that –1.5 = 3.5567 t – 4.9 t²

So we solve the quadratic equation

4.9 t² – 3.5567 t –1.5 = 0

to get t = 1.025 s

Distance traveled

d = vₓ t = 6.1604(1.025) = 6.3 m