# Finding velocity for projectile given distances and launch angle

aqryus
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.

Mentor
Homework Statement:: A cat is shot out of a cannon over a 30m high wall at an angle of 55 degrees from ground level.
Oh noes! A cat? Are you sure it wasn't a car or a rat? Poor kitty!  • PeroK, Mark44 and erobz
Homework Helper
Gold Member
I can only speak for myself when I say that I could be more helpful if understood your numbers and how you got them. I am baffled by equations like
35 = sin55 (250/cos55) - (306250/v^2cos3025)
Do you really mean cos3025? Where did you get an angle equal to 3025 in , I assume, radians?

I strongly recommend that you work out the solution algebraically without using numbers to get an expression for ##v^2## in terms of the projection angle ##\theta##, the horizontal distance ##x##, the wall height ##h## and, of course, ##g##. Then substitute the numbers. This would help us figure out (a) whether you got the wrong expression and where you went wrong or (b) whether you made a mistake in substituting the numerical values.

• berkeman and erobz
Mentor
35 = (vsin55) (250/vcos55) - (4.9) (250/vcos55)^2

canceled out the first two v

35 = sin55 (250/cos55) - (306250/v^2cos3025)
It doesn't work this way. ##(v\cos(55))^2## is ##v^2 (\cos(55))^2##, with the cosine factor commonly written as ##\cos^2(55)##. The idea is that you're supposed to calculate ##\cos(55)## and then square that number, not square 55 and take the cosine of 3025.

Really, fire a cat out of a cannon? That's weird...

• • berkeman and PeroK