- #1
salman213
- 302
- 1
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
please helppppppppppp
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
please helppppppppppp
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