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.)
- electron1Lv 71 year 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.
- oldschoolLv 71 year 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.
- Some BodyLv 71 year ago
v² = v₀² + 2a(x - x₀)
(0 m/s)² = v₀² + 2(-9.8 m/s²) (240 m - 3 m)