projectile physics?

Question

I'm trying to figure out these problems:

A model rocket carries no fuel; it is set in motion entirely by its ground-level launcher. 8.2 s after being launched, the rocket lands on the ground 415 m from the launcher.
(a) Find a component description of the rocket's initial velocity. v0x = ___m/s
v0y =____ m/s

(b) Find the speed and directional angle at which the rocket was launched. speed___ m/s
angle____ °

(c) If the launch occurs at t = 0, at what two instants t does the rocket have a speed of 47 m/s? first time ____s
second time ____s

for the first two, I tried using this formula: d=v0t + 1/2at^2. I filled everything in and got 415=v0*8.2 +1/2 * 9.8 * 8.2^2 (Im using * as multiplication signs)

I got 10.4 m/s, but the computer says I'm wrong, what am I doing wrong, and how do I do the other problems?

thanks.

Answer 1

a) well V0x should be easy: v0x = d/t = 415m/8.2s
so v0x = 50.61 m/s

since it spent half its time going up and half going down its vy just before it hit the ground would be the same as v0y which would equal a*t/2

so v0y = (9.81m/s)*(4.1s) = 40.22 m/s

b) speed = (v0x^2 + v0y^2)^1/2 = 64.65 m/s
angle = ATAN(v0y/v0x) = 38.47 degrees

c) speed = (vx^2 + vy^2)^1/2 , since there is no acceleration in the x direction vx=v0x. vy = v0y +a*t
so speed = (v0x^2 + (v0y + a*t)^2)^1/2
speed= (v0x^2 + v0y^2 +2v0y*a*t + (a*t)^2)^1/2

so (47 m/s )^2 = ((50.61 m/s)^2 + (40.22 m/s)^2 + (2*40.22*(-9.81))*t + ((-9.81 m/s/s)^2)*t^2

you solve for t and get both answers which I don't really feel like doing for you. Good luck.

Answer 2

use
vy(t)=vi*sin(th)-g*t
y(t)=vi*sin(th)*2-.5*g*t^2

x(t)=vi*cos(th)*t

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