Recent content by sermatt

After reading everything up to my post, I guess I haven't solved it after all. Agree with post #108, though.

Solved it in 20 minutes! Hell yes!

What If an astronaut who's flying perpendicular to Earth (not in orbit) starts entering the atmosphere? Will he still fry despite having a considerably lesser than orbital velocity?

So if one object is approaching me at the speed of 0.75c, and another object is approaching me at the same 0.75c but in direction opposite of the first one, the velocity with which they're closing up on each other is 0.96c? I know this has been asked a million times, but I've never paid much...

-4.9*t^2 + 36sin28*t - 7.5 = 0 b = 36sin28 t = -36sin28+-SQRT[(36sin28)^2 - 4*(-4.9)*(-7.5)] .......2*(-4.9) Best done on a calculator.

Precisely. All you have to do now is understand this concept 100% so you can find your way out of more sophisticated questions.

Remember, the vertical component does not equal 36 m/s. 36 m/s is the velocity at 28 degrees to the horizontal, so its component must be 36*sin28.

If the box is accelerating, its net force would be greater than zero. In this case, you're right, Fnet=(6.25)( 0.160 m/s^2) = 1N. Therefore, the NET force is 1N. However, net force also means that there is an unbalanced force acting on it, so 1N = Fapp - Friction (because since the object is...

Solve for horizontal displacement using the horizontal component of the initial velocity and the overall time of flight. 2.93s x 36*cos28 m/s = 93 m.

Actually, I believe you've made a mistake somewhere, so here's my solution: d = V1*t + 0.5*a*t^2 7.5 = -4.9*t^2 + 36*sin28 (to make it a beautiful quadratic equation): 0 = -4.9*t^2 + 36*sin28 - 7.5 Plug all that into a quadratic formula or a calculator, and you get two values: t = 0.523 s t =...

Does Usain Bolt play baseball? =D

The net force is the sum of all the forces acting on an object. In the case of constant velocity, the acceleration is 0, which says that there are NO unbalanced forces acting on the object. Therefore, the applied force that pushes the box must be canceled out by the force of friction of an...

Yes, it is, however, plug in 7.5 for d into the equation d = V1*t + 0.5*a*t^2 and solve for overall time the ball was in the air to correctly solve for the horizontal displacement.

Well, if you're solving for the total time the ball's been up in the air, then the grand total vertical displacement of the ball is +7.5 meters.

Where did you get -14.6 for vertical displacement?