A ball is thrown downward from the top of a 55.0 m tower with an initial speed of 15.0 m/s. Assuming negligible air resistance, what is the speed of the ball just before hitting the ground?

36.4m/s

Explanation:

According to equation of motion,

v² = u²+2as where;

v is the final velocity

u is the initial velocity = 15m/s

a is the acceleration due to gravity = +g (it is positive because the body is thrown downwards.)

g = 10m/s²

s will be the distance from the top of the tower from which it is thrown

S = 55.0m

Substituting this value to get the final velocity, we have;

v² = 15²+2(10)(55)

v² = 225+1100

v² = 1325

v =√1325

v = 36.4m/s

Therefore, the speed of the ball just before hitting the ground will be 36.4m/s

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

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.