Solve Dynamics Questions: Velocity & Height Range

[salman213]

1. http://img525.imageshack.us/img525/6376/fic11p112xe1.png
http://img525.imageshack.us/img525/9550/vqc11p112el8.png [/URL]



2. d = 1/2at^2 + v1t

v =d/t



3.
horizontal time it takes equals vertical time

Horizontal
t = d/v = 50 /(v1cos10)


Vertical
d = 1/2at^2 + v1t
-0.6 = 1/2(-9.81)(50/v1cos10)^2 + v1sin10(50 /(v1cos10)

i get v1 = 36.6 m/s

-1.6 = 1/2(-9.81)(50/v1cos10)^2 + v1sin10(50 /(v1cos10)
i get v1 = 34.8 m/s


So my range is 34.8 m/s to 36.6 m/s

for b part

dv = v1(t) - 1/2(9.81)(t^2)

maximum height occurs at half the time so i used 1/2(50/v1cos10)

dv = 34.8(1/2(50/36.64cos10) - 1/2(9.81)(1/2(50/34.8cos10))^2
i got 22.8 m

for 2nd velocity

dv = 36.64(1/2(50/36.64cos10) - 1/2(9.81)(1/2(50/36.64cos10))^2
i got 23.0 m


So my maximum height range is 22.8 m to 23.0 m

According to my online assignment this is wrong

Edit:
i just noticed for maximum height i guess I also add 2.1 m if they are measuring from the Ground so 22.8 + 2.1 = 24.9 m and 23.0 + 2.1 = 25.1 m but these values still are somehow wrong...


please helppppppppppp

 

[vector3]

$y = v_{oy} t - \frac{1}{2} g t^2 + y_o$ (1)

$y_{max}$ occurs when $y' = 0$

$$y' = v_{oy} - g t \longrightarrow t = v_{oy} / g$$

Substitute t in eq.(1)

$$y = \frac{v^2_{oy}}{g} - \frac{1}{2} g \left\{\frac{v_{oy}}{ g}\right\}^2}+ y_o$$

$$y = \frac{v^2_{oy}}{g} \left\{1 - \frac{1}{2}}\right\} + y_o$$

$$y = \frac{v^2_{oy}}{2g} + y_o$$

Using the componets of your velocities of 34.84 and 36.64 m/s should get the answer.

Should also note that Y could not exceed $25\tan{10} +2.1 \leq 6.6$ just to confirm your numbers are "in the right ball park"

 

[Nesha]

http://img301.imageshack.us/img301/2897/111tx4.th.png