Question

Two identical items, object 1 and object 2, are dropped from the top of a 50.0m50.0m building. Object 1 is dropped with an initial velocity of 0m/s0m/s, while object 2 is thrown straight downward with an initial velocity of 13.0m/s13.0m/s. What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground

Answer 1

Answer:

Δt = 1.1 s

Explanation:

Given information:

H= 50.0 m

g= 9.8 m/s²

• Object 1
• v₀ = 0

   $H = \frac{1}{2} * g* t^{2}$

Solving for t, we get:

t₁= 3.2 s

• Object 2
• v₀ = 13 m/s

We can find the final velocity for the object when it hits the ground, using the following expression:

$v_{f}^{2} - v_{o}^{2} = 2*g*H$

Solving for vf, we get:

vf = 33.9 m/s

Applying the definition of acceleration, being this acceleration the one due to gravity (g), we can write the following equation:

$v_{f} = v_{o} + g*t$

(Assuming the downward direction to be positive).

Solving for t, we get:

t₂ = 2.1 s

So the difference in time when both objects hit the ground, it's simply

Δt = t₂ - t₁ = 3.2 s - 2.1 s = 1.1 s