Time of Flight & Velocity of Projected Stone

[rachael]

A stone is prjected with a velocity of 100m/s at an elevation of 30 degrees form a tower 150m high. Find:
a. the time of flight
ans:12.(4) secs
b. the horizontal distance from the tower at which the stone strikes the ground
ans: 1.0(8) x 10 3
c. the magnitude and the direction of the velocity of the stone striking the ground.
Ans: 114m/s at 40(5) degrees below horizontal

 

[phucnv87]

http://photo-origin.tickle.com/image/69/3/7/O/69373005O558414436.jpg

 

[BobG]

A stone is prjected with a velocity of 100m/s at an elevation of 30 degrees form a tower 150m high. Find:
a. the time of flight
ans:12.(4) secs
b. the horizontal distance from the tower at which the stone strikes the ground
ans: 1.0(8) x 10 3
c. the magnitude and the direction of the velocity of the stone striking the ground.
Ans: 114m/s at 40(5) degrees below horizontal

You set up two equations, as phucnv87 did - one for the horizontal (x) and one for vertical (y). Edit: If you look at the x equation and the y equation, you might notice that they're really the same equation - in x, your start position was 0, so it was omitted; your horizontal acceleration was 0, so that part was omitted, as well.

a) You use the quadratic equation to solve the y equation for time.

Alternatively, you could use the polynomial function on your calculator, if it's permitted (you really should learn the quadratic equation, so sometimes you're not allowed to use advanced calculators). Alternatively, you could solve your quadratic equation on a slide rule (it works as well as the poynomial function on your calculator, but takes learning an entirely new skill on an instrument that's not made anymore).

b) Take your answer for 't' and plug it into the x equation to find the horizontal distance the object travels before striking the ground.

c) The only thing phucnv87 left out was how to determine your final velocity:

[tex]v_f=v_i + at[/tex]
Your initial horizontal velocity is [tex](100 m/s) (cos 30)[/tex] and your initial vertical velocity is [tex](100 m/s)(sin 30)[/tex]. You have no horizontal acceleration, but do have a vertical acceleration. Use the 't' value from a and solve for your x velocity and for your y velocity. That gives you your velocity vector in Cartesian coordinates. Use the Pythagorean Theorem to find the magnitude of the velocity vector. The tangent of your velocity vector is equal to y/x (take the arctangent of y/x to find your direction).

Alternatively, you could enter your velocity vector into your calculator as a complex number, with the x component being the real part and the y component the imaginary part. An advanced calculator will have a function that will convert a complex number from the (a + bi) form to polar form, giving the magnitude and direction with no work at all (which is why your class may not allow you to use an advanced calculator).

 

[rachael]

thank you for helping me with my work