The crew of a fairground coaster wants its 600 kg cars to pass the bottom of the first drop at 21 m/s. The cars set off from rest at the crest.
For the design estimate, the crew treats friction and air resistance as negligible.
How high must the crest be?
Well, start from the target. At the bottom, the car's kinetic store must be (0.5 × 600) × (21 × 21), which is 132300 J.
From rest at the crest, all of that must begin as GPE. Take g = 9.8 N/kg: the crest must hold (600 × 9.8) × h of gravitational energy, with the weight term coming to 5880 N.
Set the stores equal and divide: h = 132300 / 5880 = 22.5, so the crest sits 22.5 m above the bottom.
A needed launch speed runs the mirror way: the kinetic store you give at the start must cover the gravitational store at the target height.
Either direction, the state table does the bookkeeping: fill the target column first, then size the starting column to match it.
A 600 kg coaster car must pass the bottom of its first drop at 21 m/s, setting off from rest at the crest. Friction and air resistance negligible; take g = 9.8 N/kg. Calculate the crest height needed.
Write down the values:
m = 600 kg, v at the bottom = 21 m/s, v at the crest = 0 m/s, g = 9.8 N/kg, h = ?
Write down the equation:
KE_i + GPE_i = KE_f + GPE_f
The target column: KE at the bottom is (0.5 × 600) × (21 × 21), which is 132300 J, so the crest's GPE must be 132300 J.
Substitute in the values, and calculate:
h = 132300 / 5880 = 22.5
The crest must be 22.5 m above the bottom.
A 0.5 kg ball is thrown straight up and must just reach a point 2.5 m above the thrower's hand, air resistance negligible. Take g = 9.8 N/kg. Enter the launch speed needed in m/s.
A skateboarder wants to roll from rest down a smooth, friction-negligible ramp and reach 8.4 m/s at the bottom. Take g = 9.8 N/kg. Enter the drop height needed in m.
The same car, from rest at that crest, must pass a mid-ride checkpoint at 14 m/s. How far below the crest is the checkpoint?
Write down the values:
v at the crest = 0 m/s, v at the checkpoint = 14 m/s, g = 9.8 N/kg, drop = ?
Write down the equation:
KE_i + GPE_i = KE_f + GPE_f
Per kilogram, the kinetic store needed is half of (14 × 14), and the GPE given up per kilogram is 9.8 × the drop.
Substitute in the values, and calculate:
drop = 196 / 19.6 = 10
The checkpoint is 10 m below the crest.
A juggling ball must just reach a height of 1.6 m above the juggler's hand, air resistance negligible. Take g = 9.8 N/kg. Enter the launch speed needed in m/s.
A cyclist wants to coast from rest down a long, smooth driveway, rolling resistance negligible, and arrive at the bottom at 11.2 m/s. Take g = 9.8 N/kg. Enter the height the driveway must drop in m.
A 0.25 kg juggling club must just reach a point 0.9 m above the performer's hand, air resistance negligible. Take g = 9.8 N/kg. Enter the launch speed needed in m/s.
A 50 kg skateboarder crosses the top of a smooth, friction-negligible slope at 3.5 m/s and wants to pass its foot at 10.5 m/s. Take g = 9.8 N/kg. Enter the height the slope must drop in m.