You throw a 20-n rock vertically into the air from ground level. you observe that when it is a height 15.4 m above the ground, it is traveling at a speed of 24.2 m/s upward.

I believe there are two things which we asked to find here:

1. its speed just as it left the ground

2. find its maximum height

Solutions:

1. We use the formula:

ΔKE = - ΔPE

where KE is kinetic energy = ½ mv^2, and PE is potential energy = m g h, Δ = change

Therefore:

½ m (v2^2 – v1^2) = m g (h1 – h2)

at initial point, point 1: h1 = 0, v1 = ?

at final point, point 2: h2 = 15.4 m, v2 = 24.2 m/s

½ (24.2^2 – v1^2) = 9.8 (0 – 15.4)

585.64 – v1^2 = -301.84

-v1^2 = 887.48

v1 = 29.8 m/s

So the rock was travelling at 29.8 m/s as it left the ground.

2. The maximum height (hmax) reached is calculated using the formula:

v1^2 = 2 g hmax

Rewriting in terms of hmax:

hmax = v1^2 / 2 g

hmax = (29.8)^2 / (2 * 9.8)

hmax = 45.3 m

Therefore the rock reached a maximum height of 45.3 meters.

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.

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