Respuesta :
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 = [tex]\sqrt{ \frac{ 2(y_o -y)}{g} }[/tex]
calculate
t = [tex]\sqrt{ \frac{2 ( 70-50)}{9.8} }[/tex]
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
