An object is thrown upwards and after 2 seconds is 78 m above where it was initially thrown. Assume no air resistance, and determine the initial speed the object was thrown at.
To determine the initial speed at which the object was thrown, we can use the equation of motion for an object in free fall.
The equation we can use is:
h = v₀t - 0.5gt²
where:
h is the height of the object above the initial position,
v₀ is the initial velocity (what we're trying to find),
t is the time elapsed,
g is the acceleration due to gravity (9.8 m/s²).
Given the information in the question, we know:
h = 78 m
t = 2 s
g = 9.8 m/s²
Substituting the known values into the equation, we have:
78 = v₀(2) - 0.5(9.8)(2)²
Simplifying further,
78 = 2v₀ - 19.6
Rearranging the equation to isolate v₀, we get:
2v₀ = 78 + 19.6
2v₀ = 97.6
Dividing both sides by 2:
v₀ = 97.6 / 2
v₀ ≈ 48.8 m/s
Therefore, the initial speed at which the object was thrown is approximately 48.8 m/s.