If you toss a ball into the air at 22 meters per second on Earth how high will it go?

You’re in luck!

It would rise for about 19.2 meters or a little more than half a meter before falling back to the ground, though an observer on the Moon would be able to see it rise for nearly 46 meters. Why? A new measure of gravity called "g" can help us figure this out. Let's first start with a quick review of Newton's second law of motion: force equals mass times acceleration, or "F=ma." When you throw an object up vertically on Earth, the acceleration due to gravity is 9.8m/s/s--or as we know from earlier gradeschool physics, 32ft/s squared at sea level on Earth. If there were no air resistance and the ball weighed 10 grams, then it would take .45 seconds to reach the top of its arc, where it will be moving at 10m/s (or 22.7 mph, which is about how fast you'd have to throw the ball straight up to get it back down in .45 seconds). If there were no gravity, however, the ball would just keep going faster and faster as it moved away from Earth's surface. As an object reaches lower orbital speeds farther away from Earth, docking takes relatively longer--which brings us back to our question about the Moon. The gravitational force between any two objects drops off according to a very simple formula: one over the distance squared. So even though g on Earth is 9.8 m/s/s near sea level, at the height of one meter above the ground, g is approximately 9.81m/s/s--just a tiny bit less than sea level value due to the very small distance between an object and its center of mass (the Earth). On the Moon, however, gravity drops off quite rapidly because there's no atmosphere to slow down objects in low orbit around it (so you'd have to throw something really fast to keep it orbiting around), and it has almost no mass compared with Earth.

**ANSWER BY AN AI**

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2