- #1

aqryus

- 6

- 1

- Homework Statement
- A cat is shot out of a cannon over a 30m high wall at an angle of 55 degrees from ground level. With what speed does it have to be launched to make it approximately 5m lover the wall if the wall is 250m from the cannon?

- Relevant Equations
- t =d/v

d = (v) (t) + (0.5) (a) (t)^2

i used t=d/v for the x direction using the formula t=250/vcos55

and subbed that into the formula for the y direction

d = (v) (t) + (0.5) (a) (t)^2

35 = (vsin55) (250/vcos55) - (4.9) (250/vcos55)^2

canceled out the first two v

35 = sin55 (250/cos55) - (306250/v^2cos3025)

35 = 357 - (306250/v^2cos3025)

35 cos3025 (v^2) = 306250

v^2 = 306250/25cos3025

v^2 = 10669

v = 103 m/s

but the answer is 54m/s so this is way off and I'm really not sure what I'm doing wrong. I thought maybe I should use 500m for x distance because the cat reaches 35m at the wall, which is only half of the trajectory (?). Any help is appreciated thank you.

and subbed that into the formula for the y direction

d = (v) (t) + (0.5) (a) (t)^2

35 = (vsin55) (250/vcos55) - (4.9) (250/vcos55)^2

canceled out the first two v

35 = sin55 (250/cos55) - (306250/v^2cos3025)

35 = 357 - (306250/v^2cos3025)

35 cos3025 (v^2) = 306250

v^2 = 306250/25cos3025

v^2 = 10669

v = 103 m/s

but the answer is 54m/s so this is way off and I'm really not sure what I'm doing wrong. I thought maybe I should use 500m for x distance because the cat reaches 35m at the wall, which is only half of the trajectory (?). Any help is appreciated thank you.