Question

Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) With what initial velocity must an object be thrown upward (from a height of 3 meters) to reach a maximum height of 460 meters? (Round your answer to one decimal place.)

Answer 1

Answer:

Vi = 94.64 m/s

Explanation:

I order to find out the initial velocity of the object, we can use third equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration = -9.8 m/s²

h = maximum height covered by object = 460 m - 3 m = 457 m

Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)

Vi = Initial Velocity = ?

Therefore,

2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²

Vi = √8957.2 m²/s²

Vi = 94.64 m/s