Recent content by garcia1

All those variables in the exponent section should have exponents of 2 themselves.

Homework Statement A 0.21 kg rock is projected from the edge of the top of a building with an initial velocity of 7.82 m/s at an angle 56 above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 10.5 m from the base of the building. How tall is the...

24.0304m/s, by using the fact that vf = 0, and then solving for VoY and getting 6.26m/s. I then plugged this into: Y = Vy^2 - VoY^2 / (2*-9.81m/s^2 / 6) = 12.01518 Multiplying by two because I used Vf = 0 at the top of the jump, I got 24.0304m. It was right!

Homework Statement A person can jump a maximum horizontal distance (by using a 45 ◦ projectile angle) of 4 m on Earth. The acceleration of gravity is 9.8 m/s 2 . What would be his maximum range on the Moon, where the free-fall acceleration is g 6 ? Answer in units of m...

Homework Statement A ball is thrown from the top of a building upward at an angle of 35◦ to the horizontal and with an initial speed of 10 m/s. The ball is thrown at a height of 43 m above the ground. How long is the ball “in flight”? The acceleration due to gravity is 9.8 m/s2 . Answer in...

So what I did next is solve for the final velocity of part 1 by using the fact that Vo = 0m/s since the rocket starts from rest. Using the equation V = Vo + at, I got the equation V = 23*11 = 253m/s With this I placed this final velocity as the initial velocity for the next segment. I...

I got the answer right, yet I'm interested to know why the acceleration vector is not included in the equation for constant velocity in this case. For doesn't the vector in this problem lie upon the y axis, which would be affected by the force of gravity?

I don't know, this answer still came up wrong on my homework, but it seems like this is the way to do it.

So I did do that through this rationale: x = volt (since a = 0, the rest of this equation cancels out) Solving for this, x = 12*.36, I got 4.32m. From here, I subtracted this from the initial distance, 250m, to yield 245.68m. With the new distance, I calculated the acceleration by using the...

The second equation is over a. Sorry, computing error.

What I thought you meant by this was that I should set both equations equal to a common variable, and then solve for T accordingly. I tried this with the variable Vf, solving for both in this way: Eq 1: Vf = Vo +at Eq 2: Vf^2 = Vo^2 + 2ax -> Vf = rad(Vo^2 + 2ax) Then setting them equal to...

Homework Statement My problem is the last one, but it is a continuation of these first two problems, so I thought I'd include them. In Mostar, Bosnia, the ultimate test of a young man’s courage once was to jump off a 400-year-old bridge (now destroyed) into the River Neretva 27 m below...

Homework Statement A toy rocket, launched from the ground, rises vertically with an acceleration of 23 m/s2 for 11 s until its motor stops. Disregarding any air resistance, what max- imum height above the ground will the rocket achieve? The acceleration of gravity is 9.8 m/s2 . Answer...

Homework Statement A ball is thrown upward from the ground with an initial speed of 51 m/s; at the same instant, a ball is dropped from a building 38 m high. After how long will the balls be at the same height? The acceleration of gravity is 9.8 m/s2 . Answer in units of s. Homework...

I used the .36s reaction time, but i see how this is wrong now. Should I be finding the final velocity of the train when the engineer hits the breaks? Is it even necessary to find this time if I can assume that the final velocity of the whole problem would be zero, since the train would stop?