Answer:
9.39 m/s
Explanation:
Using the y-direction, we can solve for the time t it takes for the cart to reach the ground.
Assume the up direction is positive and the down direction is negative.
v₀ = 0 m/s
a = -9.8 m/s²
Δy = -50 m
t = ?
Find the constant acceleration equation that contains these four variables.
Δy = v₀t + 1/2at²
Substitute known values into this equation.
-50 = (0)t + 1/2(-9.8)t²
Multiply and simplify.
-50 = -4.9t²
Divide both sides of the equation by -4.9.
10.20408163 = t²
Square root both sides of the equation.
t = 3.194382825
Now we can use this time t and solve for v₀ in the x-direction. Time is most often our link between vertical and horizontal components of projectile motion.
List out known variables in the x-direction.
v₀ = ?
t = 3.194382825 s
a = 0 m/s²
Δx = 30 m
Find the constant acceleration equation that contains these four variables.
Δx = v₀t + 1/2at²
Substitute known values into the equation.
30 = (v₀ · 3.194382825) + 1/2(0)(3.194382825)²
Multiply and simplify.
30 = v₀ · 3.194382825
Divide both sides of the equation by 3.194382825.
v₀ = 9.391485505
The cart was rolling at a velocity of 9.39 m/s (initial velocity) when it left the ledge.