A 400g lemming waddles horizontally off a small 4.9m high cliff with a horizontal
velocity of 0.5m/s.
a. How far from the base of the cliff will the lemming land?
b. What will be the lemmings speed just before landing?
c. What is the lemmings kinetic energy just before landing?
a. (0.5 m/s)*(fall time)
b. sqrt[(g*(fall time))^2 +0.5^2]
c. (M)[gH + (0.5)^2/2]
where the "fall time" is sqrt(2H/g)
H = 4.9 m
g = 9.8 m/s^2
M = 0.4 kg
Crank it out.
A ball rolls horizontally of an 85 m cliff with a velocity of 11 m/s. How far from the base of the cliff will it land?
To solve this problem, we can use the equations of motion for projectiles. Let's break it down step by step:
a. How far from the base of the cliff will the lemming land?
We are given the vertical height (h) of the cliff, which is 4.9m, and the horizontal velocity (v_x) of the lemming, which is 0.5m/s.
First, let's find the time it takes for the lemming to hit the ground using the equation:
h = (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time. Rearranging the equation to solve for t:
t = sqrt(2h / g)
Substituting the given values:
t = sqrt(2 * 4.9 / 9.8) = sqrt(0.98) = 0.99 seconds (approximately)
To find how far from the base of the cliff the lemming will land, we can use the equation:
distance = velocity * time
Since the lemming's horizontal velocity is 0.5 m/s, and the time is 0.99 seconds, we can substitute the values into the equation:
distance = 0.5 * 0.99 = 0.495 meters (approximately)
Therefore, the lemming will land approximately 0.495 meters from the base of the cliff.
b. What will be the lemming's speed just before landing?
To find the lemming's speed just before landing, we can use the equation:
v = sqrt(v_x^2 + v_y^2)
where v is the total velocity, v_x is the horizontal velocity, and v_y is the vertical velocity.
The horizontal velocity is given as 0.5 m/s, and the vertical velocity can be calculated using the equation:
v_y = g * t
Substituting the values:
v_y = 9.8 * 0.99 = 9.702 m/s (approximately)
Calculating the total velocity:
v = sqrt(0.5^2 + 9.702^2) = sqrt(0.25 + 94.114404) = sqrt(94.364404) = 9.715 m/s (approximately)
Therefore, the lemming's speed just before landing is approximately 9.715 m/s.
c. What is the lemming's kinetic energy just before landing?
The kinetic energy (KE) of an object is given by the equation:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
The mass of the lemming is given as 400g, which we need to convert to kilograms:
m = 400g / 1000 = 0.4kg
Substituting the values:
KE = (1/2) * 0.4 * 9.715^2 = 21.9188 Joules (approximately)
Therefore, the lemming's kinetic energy just before landing is approximately 21.9188 Joules.