The ride in a giant roller coaster includes a vertical drop of 115 m. If the coaster has a speed of 5.5 m/s at the top of the drop, find the speed of the riders at the bottom.
To find the speed of the riders at the bottom of the drop, we can use the principle of conservation of energy. The total mechanical energy of the riders at the top of the drop is equal to the total mechanical energy at the bottom of the drop.
At the top of the drop, the riders have gravitational potential energy and kinetic energy, which is given by:
E_top = m * g * h + (1/2) * m * v_top^2
where:
m is the mass of the riders
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height of the drop (115 m)
v_top is the speed of the riders at the top of the drop (5.5 m/s)
At the bottom of the drop, the riders only have kinetic energy, given by:
E_bottom = (1/2) * m * v_bottom^2
where:
v_bottom is the speed of the riders at the bottom of the drop.
Since the total mechanical energy is conserved, we can equate E_top to E_bottom:
m * g * h + (1/2) * m * v_top^2 = (1/2) * m * v_bottom^2
We can now solve for v_bottom:
v_bottom^2 = (2 * (m * g * h + (1/2) * m * v_top^2)) / m
v_bottom^2 = 2 * g * h + v_top^2
Taking the square root of both sides, we have:
v_bottom = sqrt(2 * g * h + v_top^2)
Now we can substitute the values:
v_bottom = sqrt(2 * 9.8 m/s^2 * 115 m + 5.5 m/s^2)^2
v_bottom = sqrt(2 * 9.8 * 115 + 5.5^2)
v_bottom = sqrt(2254)
v_bottom ≈ 47.50 m/s
Therefore, the speed of the riders at the bottom of the drop is approximately 47.50 m/s.