Question

is
A 0.155 kg arrow is shot upward
at 31.4 m/s. What is its kinetic
energy (KE) when it is 30.0 m
above the ground?

Answer 1

Kinetic energy = 29.912 J

Further explanation

Given

mass = 0.155 kg

vo=initial velocity = 31.4 m/s

h/d=30 m

Required

KE=kinetic energy

Solution

vt²=vo²-2ad⇒vt=final velocity

vt²=31.4²-2.10.30⇒g=10 m/s²

vt²=985.96-600

vt²=385.96 m/s

Kinetic  energy (KE)

$\tt KE=\dfrac{1}{2}mv^2\\\\KE=\dfrac{1}{2}\times 0.155.\times 385.96\\\\KE=29.912~J$

Answer 2

Answer:

The answer is 30.8 (J)

Explanation:

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