# May I please get help with physics?

A softball coach throws a 1 kg ball horizontally to her player with an initial velocity of 40 m/s. The ball leaves her hand at 6 m high, and her catcher is 20 meters away.

1. What is the initial horizontal velocity, vxi, in m/s?

2. What is the initial vertical velocity, vyi, in m/s? (+ is for up, - is for down)

3. What is the horizontal acceleration, ax, in m/s2?

22. The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km. A 750 kg satellite orbits 35,800 km above Earth in a perfectly circular orbit. What is the gravitational force on the satellite due to Earth?

### 2 Answers

- ♥Astrid♥Lv 61 month agoFavourite answer
(1) The initial horizontal velocity Vxi,is given as 40 m/s.

(2) The initial vertical velocity Vyi is given as 0 m/s

(3) Once the ball leaves the hand the only force is gravity in the vertical direction, thus the horizontal acceleration is zero.

(22)

F = GMm / (r + h)²

M = 5.972 * 10^24 kg

m = 750 kg

r = 6,371 km

h = 35,800 km

G = 6.67408 × 10^-11 N•m²/kg²

F = (6.67408 × 10^-11)(5.972 * 10^24)(750) / (6,371,000 + 35,800,000)²

F = 168 N

- oubaasLv 71 month ago
A softball coach throws a 1 kg ball horizontally to her player with an initial velocity of 40 m/s. The ball leaves her hand at 6 m high, and her catcher is 20 meters away.

1. What is the initial horizontal velocity, vxi, in m/s?

Vxi = 40 m/sec

2. What is the initial vertical velocity, vyi, in m/s? (+ is for up, - is for down)

Vyi = zero m/sec

3. What is the horizontal acceleration, ax, in m/s2?

ax = 0 m/sec^2, since no other force but gravity act on the ball

22.

The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km. A 750 kg satellite orbits 35,800 km above Earth in a perfectly circular orbit. What is the gravitational force on the satellite due to Earth?

g = Me*G/(re+h)^2

g = 5.972*10^24*6.674*10^-11/(6.371+35.8)^2*10^12) = 0.2241 m/sec^2

Fg = m*g = 0.2241*750 = 168.1 N ...(168 with just 3 significant figures)