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# Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.)?

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 240 meters? (Round your answer to one decimal place.)

### 3 Answers

- electron1Lv 710 months ago
At the maximum height, the object’s velocity is 0 m/s. Let’s use the following equation to determine its initial velocity.

vf^2 = vi^2 + 2 * a * d

d = final height – initial height = 240 – 3 = 237 meters

0 = vi^2 + 2 * -9.8 * 237

vi = √4,645.2

The initial velocity is approximately 68 m/s.

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- oldschoolLv 710 months ago
You could use conservation of energy (neglecting air resistance)

mg(240-3) = ½mVi²

g(240-3) = ½Vi²

2*g(240-3) = Vi² = 68.2² so the launch velocity is about 68m/s to reach a height of 240m from a 3m launch height.

OR we could do it this way

Height(t) = ho + Vyi*t - ½*g*t²

But Vyi - gt = 0 at max height so t = Vyi/g at h(t) = 240m

240 = 3 + Vyi²/g - ½*g*Vyi²/g² -----> 237 = Vyi²/2g ----->

19.6*237 = Vyi² = 68.2m/s same answer.

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- Some BodyLv 710 months ago
v² = v₀² + 2a(x - x₀)

(0 m/s)² = v₀² + 2(-9.8 m/s²) (240 m - 3 m)

v₀ =

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- Some BodyLv 710 months agoReport
Correct.

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