Question

What minimum speed must the block have at the base of the 70 m hill to pass over the pit at the far (right-hand) side of that hill and reach the other side (at a 50 m level)

Answer 1

Answer:

initial velocity is v = 4.95 m / s

Explanation:

To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.

Initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

          y-y₀ = 0 -1/2 g t²

          t = $\sqrt{ \frac{ 2(y_o -y)}{g} }$

calculate

          t = $\sqrt{ \frac{2 ( 70-50)}{9.8} }$

          t = 2.02 s

with this time we can substitute in the horizontal displacement equation

          x = v₀ₓ t

          v₀ₓ = x / t

suppose that the distance between the two points is x = 10 m

          v₀ₓ = 10 / 2.02

          v₀ₓ = 4.95 m / s

initial velocity is v = 4.95 m / s